Do higher dimensions have axes? [closed]Orthogonal planes in n-dimensionsWhat makes us say that Pythagoras theorem can be used in higher dimensions too?Pairwise tangent circles radical axesI have a hard time understanding this simple theorem: “If two lines intersect, then exactly one plane contains the lines.”What are the higher dimensional analogues of left and right handedness?What's new in higher dimensions?Retroreflectors in higher dimensions?Euclidean & Cartesian Geometry in infinite dimensionsIs there a higher dimension equivalent of lattices with lines instead of points?
Ball-passing game with a surprise ending
Divorced woman remarrying her ex-husband
Implement the Max-Pooling operation from Convolutional Neural Networks
Is it possible to keep cat litter on balcony during winter (down to -10°C)
How to draw the following image in TikZ?
Mechanism for the reaction between trisilylamine and hydrogen chloride
Explanatory vs Non-explanatory Proofs
Extra long braces with beginarray[t]
Which modifier shown in the D&D Beyond character sheet do I add to attack rolls with my longbow?
What would happen to the world if all lightning stopped?
Python Sound visualizer - appreciate feedback
Why did Leia not want to tell Han about Luke being her twin brother?
What is the purpose of this single box found in Rome B&B room?
My name causes an issue with any booking! (names end with MR and MRS)
Was the whistle-blower's (12 Aug 2019) complaint deemed credible?
Is it possible to write a short story in 500 words?
Is it sportsmanlike to waste opponents' time by giving check at the end of the game?
Calculate the sum of interior angles of a polygon
Will a nuclear country use nuclear weapons if attacked by conventional means by another nuclear country?
Is a vector space naturally isomorphic to its dual?
Why do airports in the UK have so few runways?
What is the difference between turbojet and turbofan engines?
Can "marriage" be used as a verb?
If a middle class person is expecting a down turn in the market, what percentage of cash should they hold?
Do higher dimensions have axes? [closed]
Orthogonal planes in n-dimensionsWhat makes us say that Pythagoras theorem can be used in higher dimensions too?Pairwise tangent circles radical axesI have a hard time understanding this simple theorem: “If two lines intersect, then exactly one plane contains the lines.”What are the higher dimensional analogues of left and right handedness?What's new in higher dimensions?Retroreflectors in higher dimensions?Euclidean & Cartesian Geometry in infinite dimensionsIs there a higher dimension equivalent of lattices with lines instead of points?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;
$begingroup$
I understand that the three dimensions each have their own axes, for lines, planes and volumes, and that 4 dimension has an axis but it is complicated and hard to determine, but from what I understand is that it is there.
So I was wondering, disregarding their complexity, do Higher-Dimensions such as 5D to 10/11D have axes?
geometry
edited Jul 19 at 1:57
Tanner Swett
6,52322 silver badges42 bronze badges
asked Jul 19 at 1:46
C. JordanC. Jordan
1021 silver badge7 bronze badges
$endgroup$
closed as unclear what you're asking by Matthew Towers, Paul Frost, José Carlos Santos, YuiTo Cheng, Daniele Tampieri Jul 21 at 12:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment
|
$begingroup$
I understand that the three dimensions each have their own axes, for lines, planes and volumes, and that 4 dimension has an axis but it is complicated and hard to determine, but from what I understand is that it is there.
So I was wondering, disregarding their complexity, do Higher-Dimensions such as 5D to 10/11D have axes?
geometry
edited Jul 19 at 1:57
Tanner Swett
6,52322 silver badges42 bronze badges
asked Jul 19 at 1:46
C. JordanC. Jordan
1021 silver badge7 bronze badges
$endgroup$
closed as unclear what you're asking by Matthew Towers, Paul Frost, José Carlos Santos, YuiTo Cheng, Daniele Tampieri Jul 21 at 12:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
what do you mean by an axis for volume?
$endgroup$
– Mirko
Jul 19 at 3:14
$begingroup$
@Mirko as in three dimensional space
$endgroup$
– C. Jordan
Jul 19 at 3:36
15
$begingroup$
Heh heh. I am thinking about a higher-dimensional hero swinging an ax at the head of a Killing Form to defeat the Lie Algebra.
$endgroup$
– puppetsock
Jul 19 at 13:42
4
$begingroup$
I'd say that's pretty much the definition of a dimension - think of it as another set of directions you can go. 1D lets you go only left and right, 2D adds forward and backward, 3D adds up and down, etc. Each dimension is just like the previous, only with another axis at a right angle to all the previous axes. (Obviously this becomes difficult to visualize past 3, but the math doesn't really care about that, it works pretty much the same for any number of dimensions.)
$endgroup$
– Darrel Hoffman
Jul 19 at 14:29
$begingroup$
@DarrelHoffman I don't see what makes it so hard to visualize. I just use a trick that my high school pre-calculus teacher taught me ages ago: if you want to visualize a four dimensional space, first just visualize an $n$-dimensional space, then take $n=4$. What could be simpler? :P
$endgroup$
– Xander Henderson
Jul 20 at 13:09
add a comment
|
$begingroup$
I understand that the three dimensions each have their own axes, for lines, planes and volumes, and that 4 dimension has an axis but it is complicated and hard to determine, but from what I understand is that it is there.
So I was wondering, disregarding their complexity, do Higher-Dimensions such as 5D to 10/11D have axes?
geometry
edited Jul 19 at 1:57
Tanner Swett
6,52322 silver badges42 bronze badges
asked Jul 19 at 1:46
C. JordanC. Jordan
1021 silver badge7 bronze badges
$endgroup$
I understand that the three dimensions each have their own axes, for lines, planes and volumes, and that 4 dimension has an axis but it is complicated and hard to determine, but from what I understand is that it is there.
So I was wondering, disregarding their complexity, do Higher-Dimensions such as 5D to 10/11D have axes?
geometry
geometry
edited Jul 19 at 1:57
Tanner Swett
6,52322 silver badges42 bronze badges
asked Jul 19 at 1:46
C. JordanC. Jordan
1021 silver badge7 bronze badges
edited Jul 19 at 1:57
Tanner Swett
6,52322 silver badges42 bronze badges
asked Jul 19 at 1:46
C. JordanC. Jordan
1021 silver badge7 bronze badges
edited Jul 19 at 1:57
Tanner Swett
6,52322 silver badges42 bronze badges
edited Jul 19 at 1:57
Tanner Swett
6,52322 silver badges42 bronze badges
edited Jul 19 at 1:57
Tanner Swett
6,52322 silver badges42 bronze badges
6,52322 silver badges42 bronze badges
asked Jul 19 at 1:46
C. JordanC. Jordan
1021 silver badge7 bronze badges
asked Jul 19 at 1:46
C. JordanC. Jordan
1021 silver badge7 bronze badges
asked Jul 19 at 1:46
C. JordanC. Jordan
1021 silver badge7 bronze badges
1021 silver badge7 bronze badges
closed as unclear what you're asking by Matthew Towers, Paul Frost, José Carlos Santos, YuiTo Cheng, Daniele Tampieri Jul 21 at 12:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Matthew Towers, Paul Frost, José Carlos Santos, YuiTo Cheng, Daniele Tampieri Jul 21 at 12:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Matthew Towers, Paul Frost, José Carlos Santos, YuiTo Cheng, Daniele Tampieri Jul 21 at 12:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
what do you mean by an axis for volume?
$endgroup$
– Mirko
Jul 19 at 3:14
$begingroup$
@Mirko as in three dimensional space
$endgroup$
– C. Jordan
Jul 19 at 3:36
15
$begingroup$
Heh heh. I am thinking about a higher-dimensional hero swinging an ax at the head of a Killing Form to defeat the Lie Algebra.
$endgroup$
– puppetsock
Jul 19 at 13:42
4
$begingroup$
I'd say that's pretty much the definition of a dimension - think of it as another set of directions you can go. 1D lets you go only left and right, 2D adds forward and backward, 3D adds up and down, etc. Each dimension is just like the previous, only with another axis at a right angle to all the previous axes. (Obviously this becomes difficult to visualize past 3, but the math doesn't really care about that, it works pretty much the same for any number of dimensions.)
$endgroup$
– Darrel Hoffman
Jul 19 at 14:29
$begingroup$
@DarrelHoffman I don't see what makes it so hard to visualize. I just use a trick that my high school pre-calculus teacher taught me ages ago: if you want to visualize a four dimensional space, first just visualize an $n$-dimensional space, then take $n=4$. What could be simpler? :P
$endgroup$
– Xander Henderson
Jul 20 at 13:09
add a comment
|
3
$begingroup$
what do you mean by an axis for volume?
$endgroup$
– Mirko
Jul 19 at 3:14
$begingroup$
@Mirko as in three dimensional space
$endgroup$
– C. Jordan
Jul 19 at 3:36
15
$begingroup$
Heh heh. I am thinking about a higher-dimensional hero swinging an ax at the head of a Killing Form to defeat the Lie Algebra.
$endgroup$
– puppetsock
Jul 19 at 13:42
4
$begingroup$
I'd say that's pretty much the definition of a dimension - think of it as another set of directions you can go. 1D lets you go only left and right, 2D adds forward and backward, 3D adds up and down, etc. Each dimension is just like the previous, only with another axis at a right angle to all the previous axes. (Obviously this becomes difficult to visualize past 3, but the math doesn't really care about that, it works pretty much the same for any number of dimensions.)
$endgroup$
– Darrel Hoffman
Jul 19 at 14:29
$begingroup$
@DarrelHoffman I don't see what makes it so hard to visualize. I just use a trick that my high school pre-calculus teacher taught me ages ago: if you want to visualize a four dimensional space, first just visualize an $n$-dimensional space, then take $n=4$. What could be simpler? :P
$endgroup$
– Xander Henderson
Jul 20 at 13:09
3
3
$begingroup$
what do you mean by an axis for volume?
$endgroup$
– Mirko
Jul 19 at 3:14
$begingroup$
what do you mean by an axis for volume?
$endgroup$
– Mirko
Jul 19 at 3:14
$begingroup$
@Mirko as in three dimensional space
$endgroup$
– C. Jordan
Jul 19 at 3:36
$begingroup$
@Mirko as in three dimensional space
$endgroup$
– C. Jordan
Jul 19 at 3:36
15
15
$begingroup$
Heh heh. I am thinking about a higher-dimensional hero swinging an ax at the head of a Killing Form to defeat the Lie Algebra.
$endgroup$
– puppetsock
Jul 19 at 13:42
$begingroup$
Heh heh. I am thinking about a higher-dimensional hero swinging an ax at the head of a Killing Form to defeat the Lie Algebra.
$endgroup$
– puppetsock
Jul 19 at 13:42
4
4
$begingroup$
I'd say that's pretty much the definition of a dimension - think of it as another set of directions you can go. 1D lets you go only left and right, 2D adds forward and backward, 3D adds up and down, etc. Each dimension is just like the previous, only with another axis at a right angle to all the previous axes. (Obviously this becomes difficult to visualize past 3, but the math doesn't really care about that, it works pretty much the same for any number of dimensions.)
$endgroup$
– Darrel Hoffman
Jul 19 at 14:29
$begingroup$
I'd say that's pretty much the definition of a dimension - think of it as another set of directions you can go. 1D lets you go only left and right, 2D adds forward and backward, 3D adds up and down, etc. Each dimension is just like the previous, only with another axis at a right angle to all the previous axes. (Obviously this becomes difficult to visualize past 3, but the math doesn't really care about that, it works pretty much the same for any number of dimensions.)
$endgroup$
– Darrel Hoffman
Jul 19 at 14:29
$begingroup$
@DarrelHoffman I don't see what makes it so hard to visualize. I just use a trick that my high school pre-calculus teacher taught me ages ago: if you want to visualize a four dimensional space, first just visualize an $n$-dimensional space, then take $n=4$. What could be simpler? :P
$endgroup$
– Xander Henderson
Jul 20 at 13:09
$begingroup$
@DarrelHoffman I don't see what makes it so hard to visualize. I just use a trick that my high school pre-calculus teacher taught me ages ago: if you want to visualize a four dimensional space, first just visualize an $n$-dimensional space, then take $n=4$. What could be simpler? :P
$endgroup$
– Xander Henderson
Jul 20 at 13:09
add a comment
|
7 Answers
7
active
oldest
votes
$begingroup$
They sure do. In three dimensions,
- the $x$-axis is the set of all points where both coordinates besides $x$ are zero,
- the $y$-axis is the set of all points where both coordinates besides $y$ are zero, and
- the $z$-axis is the set of all points where both coordinates besides $z$ are zero.
If you're looking at a $26$-dimensional space with coordinates $(a, b, c, ldots, z)$, then
- the $a$-axis is the set of all points where all coordinates besides $a$ are zero, and
- the $b$-axis is the set of all points where all coordinates besides $b$ are zero, and
- the $c$-axis is the set of all points where all coordinates besides $c$ are zero, and
- ...
- the $z$-axis is the set of all points where all coordinates besides $z$ are zero.
answered Jul 19 at 1:55
Tanner SwettTanner Swett
6,52322 silver badges42 bronze badges
$endgroup$
$begingroup$
Thank you, this will help, I am wondering if this can be somewhat similarly said for physics or theoretical physics as well
$endgroup$
– C. Jordan
Jul 19 at 13:51
11
$begingroup$
I feel like even though this answer technically provides the direct answer to the OP's question, since the question itself is aimed at better understanding higher dimensional spaces, this answer could be improved by mentioning that while those space do have axes, not everything about axes in 3d applies to higher dimensional spaces. Such as that in dimensions higher than 3, a single axis no longer uniquely determines a rotation.
$endgroup$
– Shufflepants
Jul 19 at 14:59
14
$begingroup$
@Shufflepants - The OP appears to be speaking strictly about coordinate axes. So mixing in the axis of a rotation (which has nothing to do with coordinate axes) is just sowing confusion. And it is only in 3D that you can associate rotations with an axis. In 2D, they are associated with points, not lines, and in 1D rotations don't even exist. More generally, a rotation is associated with an affine space of codimension 2.
$endgroup$
– Paul Sinclair
Jul 19 at 16:58
$begingroup$
What about the vector space $mathbbC^2$ over $mathbbC$?
$endgroup$
– Paracosmiste
Jul 19 at 21:35
add a comment
|
$begingroup$
Sure. In $n$ dimensions, the $k^textth$ axis, for some $1 leq k leq n$, is the span of the vector whose $k^textth$ component is $1$ and all other components $0$. The problem is that there's not a particularly satisfying geometric visualization of these axes and one largely has to settle for this algebraic abstraction.
answered Jul 19 at 1:51
Kaj HansenKaj Hansen
29k4 gold badges41 silver badges85 bronze badges
$endgroup$
add a comment
|
$begingroup$
The first important term here is that of a vector space. This is a general term; vector spaces can be far more than simply some $mathbbR^n$. Polynomials, for example, form a vector space of infinite dimension.
Now you need to understand what a dimension is. If you have a vector in three dimensions, you write that vector like $(4,1,7)$. Those coordinates make sense in the context of a base. A base is a set of linearly independent vectors, which has the maximal possible size for that vector space. This size is the dimension. A three dimensional space is defined by having maximally three linear independent vectors in a set.
Edit: On the recommendation of Kaj Hansen, let me at this point make clear that a base of a given vector space has always the same amount of vectors. This is important, because this defines the dimension of the vector space.
The coordinates are now in relation to your base. If your base is $b_1, b_2, b_3$, the vector $(4,1,7)$ is the vector you get by the linear combination $4b_1 + 1b_2 + 7b_3$.
Now realize that this base can change. You can take three linear independent vectors different from your current base and make them the new base, resulting in all vectors getting different coordinates.
To give you an example when this is used, consider a 3D image in computer graphics that depicts one of those wooden puppets that artists use. There is a global coordinate system for the whole image, and the puppet has coordinates within that. But the limps have local coordinate systems which are in relation to the joints of the puppet. Hope this helps you.
Now, for the axes, an axis is simply the line created when you follow the direction of a base vector. Nothing more. They are arbitrary, and they do not even have to be orthogonal to each other. They are not a fundamental principle of a space. Coordinates are just numbers how we model it.
Another point of view, let's say that you are to determine the axes of reality, of the three-dimensional space we live in. In which direction would you point the first axis? Arbitrary. You decide on some direction for the first an then for the others, and then you write things in relation to those (or, better to say, the base vectors, as you also need a length). There is not the one objective way to create your axes.
Edit: Inserted from the comment of Kaj Hansen, note that usually one uses the standard base $e_k^n_k=1$ for $mathbbR^n$, which consists of vectors $k_i$ that are all zero except for the $k^th$ component which is one. With those, a coordinate evaluates to a vector with the same numbers. Now, I brought this here and not above for a reason, and that is because we just talked about the "axes of reality", which do not exist. In the real world, there is no objective coordinate (1, 0, 0). For that reason, even the standard base is not "objective" or anything, but bound to some definitions we made. When we compute things in math, (1, 0, 0) is an objective thing, though, because in math, we live only in definitions and axioms.
With more dimensions, it is not different. You have a base and you work with it, and you can totally change the base.
Things might get difficult when you talk about physical spaces, though. Time behaves different as a dimension, and we are now in a Minkowski space, and things are getting complicated.
If you are interested in examples for vector spaces which aren't R^n, like the polynomial vector spaces or that of other functions, I can explain that too, but as this is long enough already, I'll stop for now.
answered Jul 19 at 10:25
AziuthAziuth
1547 bronze badges
$endgroup$
$begingroup$
That is alright, I think I get the picture now, sounds like dimensions do have at least one axis for each one, even for 4 dimensional and beyond
$endgroup$
– C. Jordan
Jul 19 at 13:53
$begingroup$
I salute you for posting this. The word "dimension" is probably the most misunderstood mathematical term that's entered lay vocabulary, and ultimately something like this is what's needed to explain it properly.
$endgroup$
– Kaj Hansen
Jul 20 at 1:39
$begingroup$
Two recommended additions: explicitly point out the facts that 1) any two bases have the same number of vectors; and 2) we generally, unless otherwise stated, use the standard basis $mathbfe_k_k=1^n$ for $mathbbR^n$, where $mathbfe_k$ is the vector whose $k^textth$ component is $1$ and all others $0$. Hence, writing $langle 2, 5, 3 rangle$ is implicitly saying $2 langle 1, 0, 0 rangle + 5 langle 0, 1, 0 rangle + 3 langle 0, 0, 1 rangle$.
$endgroup$
– Kaj Hansen
Jul 20 at 5:03
$begingroup$
@KajHansen Inserted your recommendations. Continued on your second one with some more "philosophical" aspects of math, thought they belong nicely in a post about the basic idea of a dimension.
$endgroup$
– Aziuth
Jul 20 at 8:37
add a comment
|
$begingroup$
In three dimensions the three axes each perpendicular to the other two tell you that you identify a point by giving its coordinates $(x,y,z)$.
The three axes are the lists of coordinates that have just one entry that's not $0$.
Then you can do geometry by doing algebra with the coordinates.
For example, the eight points
(0, 0, 0) (1, 1, 0)
(1, 0, 0) (1, 0, 1)
(0, 1, 0) (0, 1, 1)
(0, 0, 1) (1, 1, 1)
are the vertices of a cube. (Those are all the lists of coordinates you can make using just $0$ and $1$.)
If you just allow yourself more coordinates you can study higher dimensions. So the $16$ ways to make all the lists of four coordinates each of which is just $0$ and $1$ are the vertices of the analogue of a cube in four dimensions. You can study its properties. Actually "seeing" it as a geometric object is pretty much beyond our human abilities. Here's the best we can do: https://en.wikipedia.org/wiki/Tesseract .
Clearly there's no need to stop at four coordinates/dimensions.
answered Jul 19 at 2:04
Ethan BolkerEthan Bolker
58.6k5 gold badges64 silver badges140 bronze badges
$endgroup$
add a comment
|
$begingroup$
Yes. Forget everything you previously saw with the cartesian coordinate system, etc.
Now, imagine a ball of yarn with a bunch of needles sticking out of it. Those dimensions each have a needle (an axis for that dimension). Now, when you have a bunch of dimensions, each thing you add to your system needs to have a thread of yarn that touches each needle somewhere along its length.
Each thread is a vector, or the position of that thing you added to your system.
answered Jul 19 at 20:04
JakeJJakeJ
1291 bronze badge
$endgroup$
add a comment
|
$begingroup$
Each dimension will, by definition, have an axis that is 90 degrees askew to all other dimensions. Since our everyday world is 3-dimensional, this may be difficult to envision/imagine for axes beyond the basic x-y-z, but mathematically this is how it works.
answered Jul 19 at 17:10
EJSawyerEJSawyer
191 bronze badge
$endgroup$
$begingroup$
Use MathJax please
$endgroup$
– Aqua
Jul 19 at 17:39
$begingroup$
Thank you very much, this is what I wish to hear
$endgroup$
– C. Jordan
Jul 19 at 18:27
$begingroup$
From where do the 90 degrees come from? Any subspace is a proper space, no matter how it is angled to some axis. Plus, you make it sound as if a "dimension" is something particular, rather than a number of independencies.
$endgroup$
– Aziuth
Jul 20 at 8:41
add a comment
|
$begingroup$
Proof: Suppose $n$ is the highest dimension for which we have axes. Let $(x_1,x_2,ldots, x_n)$ be any arbitrary point in this dimension. By assumption there are no axes in $n+1$ dimension, so there must be infinitely many points with the exact same coordinates $(x_1,x_2,ldots, x_n,666)$ i.e. unique points in $n+1$ dimension do not exist which is a contradiction to the definition of a dimension.
answered Jul 20 at 1:53
Nilotpal Kanti SinhaNilotpal Kanti Sinha
7,7802 gold badges19 silver badges52 bronze badges
$endgroup$
add a comment
|
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
They sure do. In three dimensions,
- the $x$-axis is the set of all points where both coordinates besides $x$ are zero,
- the $y$-axis is the set of all points where both coordinates besides $y$ are zero, and
- the $z$-axis is the set of all points where both coordinates besides $z$ are zero.
If you're looking at a $26$-dimensional space with coordinates $(a, b, c, ldots, z)$, then
- the $a$-axis is the set of all points where all coordinates besides $a$ are zero, and
- the $b$-axis is the set of all points where all coordinates besides $b$ are zero, and
- the $c$-axis is the set of all points where all coordinates besides $c$ are zero, and
- ...
- the $z$-axis is the set of all points where all coordinates besides $z$ are zero.
answered Jul 19 at 1:55
Tanner SwettTanner Swett
6,52322 silver badges42 bronze badges
$endgroup$
$begingroup$
Thank you, this will help, I am wondering if this can be somewhat similarly said for physics or theoretical physics as well
$endgroup$
– C. Jordan
Jul 19 at 13:51
11
$begingroup$
I feel like even though this answer technically provides the direct answer to the OP's question, since the question itself is aimed at better understanding higher dimensional spaces, this answer could be improved by mentioning that while those space do have axes, not everything about axes in 3d applies to higher dimensional spaces. Such as that in dimensions higher than 3, a single axis no longer uniquely determines a rotation.
$endgroup$
– Shufflepants
Jul 19 at 14:59
14
$begingroup$
@Shufflepants - The OP appears to be speaking strictly about coordinate axes. So mixing in the axis of a rotation (which has nothing to do with coordinate axes) is just sowing confusion. And it is only in 3D that you can associate rotations with an axis. In 2D, they are associated with points, not lines, and in 1D rotations don't even exist. More generally, a rotation is associated with an affine space of codimension 2.
$endgroup$
– Paul Sinclair
Jul 19 at 16:58
$begingroup$
What about the vector space $mathbbC^2$ over $mathbbC$?
$endgroup$
– Paracosmiste
Jul 19 at 21:35
add a comment
|
$begingroup$
They sure do. In three dimensions,
- the $x$-axis is the set of all points where both coordinates besides $x$ are zero,
- the $y$-axis is the set of all points where both coordinates besides $y$ are zero, and
- the $z$-axis is the set of all points where both coordinates besides $z$ are zero.
If you're looking at a $26$-dimensional space with coordinates $(a, b, c, ldots, z)$, then
- the $a$-axis is the set of all points where all coordinates besides $a$ are zero, and
- the $b$-axis is the set of all points where all coordinates besides $b$ are zero, and
- the $c$-axis is the set of all points where all coordinates besides $c$ are zero, and
- ...
- the $z$-axis is the set of all points where all coordinates besides $z$ are zero.
answered Jul 19 at 1:55
Tanner SwettTanner Swett
6,52322 silver badges42 bronze badges
$endgroup$
$begingroup$
Thank you, this will help, I am wondering if this can be somewhat similarly said for physics or theoretical physics as well
$endgroup$
– C. Jordan
Jul 19 at 13:51
11
$begingroup$
I feel like even though this answer technically provides the direct answer to the OP's question, since the question itself is aimed at better understanding higher dimensional spaces, this answer could be improved by mentioning that while those space do have axes, not everything about axes in 3d applies to higher dimensional spaces. Such as that in dimensions higher than 3, a single axis no longer uniquely determines a rotation.
$endgroup$
– Shufflepants
Jul 19 at 14:59
14
$begingroup$
@Shufflepants - The OP appears to be speaking strictly about coordinate axes. So mixing in the axis of a rotation (which has nothing to do with coordinate axes) is just sowing confusion. And it is only in 3D that you can associate rotations with an axis. In 2D, they are associated with points, not lines, and in 1D rotations don't even exist. More generally, a rotation is associated with an affine space of codimension 2.
$endgroup$
– Paul Sinclair
Jul 19 at 16:58
$begingroup$
What about the vector space $mathbbC^2$ over $mathbbC$?
$endgroup$
– Paracosmiste
Jul 19 at 21:35
add a comment
|
$begingroup$
They sure do. In three dimensions,
- the $x$-axis is the set of all points where both coordinates besides $x$ are zero,
- the $y$-axis is the set of all points where both coordinates besides $y$ are zero, and
- the $z$-axis is the set of all points where both coordinates besides $z$ are zero.
If you're looking at a $26$-dimensional space with coordinates $(a, b, c, ldots, z)$, then
- the $a$-axis is the set of all points where all coordinates besides $a$ are zero, and
- the $b$-axis is the set of all points where all coordinates besides $b$ are zero, and
- the $c$-axis is the set of all points where all coordinates besides $c$ are zero, and
- ...
- the $z$-axis is the set of all points where all coordinates besides $z$ are zero.
answered Jul 19 at 1:55
Tanner SwettTanner Swett
6,52322 silver badges42 bronze badges
$endgroup$
They sure do. In three dimensions,
- the $x$-axis is the set of all points where both coordinates besides $x$ are zero,
- the $y$-axis is the set of all points where both coordinates besides $y$ are zero, and
- the $z$-axis is the set of all points where both coordinates besides $z$ are zero.
If you're looking at a $26$-dimensional space with coordinates $(a, b, c, ldots, z)$, then
- the $a$-axis is the set of all points where all coordinates besides $a$ are zero, and
- the $b$-axis is the set of all points where all coordinates besides $b$ are zero, and
- the $c$-axis is the set of all points where all coordinates besides $c$ are zero, and
- ...
- the $z$-axis is the set of all points where all coordinates besides $z$ are zero.
answered Jul 19 at 1:55
Tanner SwettTanner Swett
6,52322 silver badges42 bronze badges
answered Jul 19 at 1:55
Tanner SwettTanner Swett
6,52322 silver badges42 bronze badges
answered Jul 19 at 1:55
Tanner SwettTanner Swett
6,52322 silver badges42 bronze badges
answered Jul 19 at 1:55
Tanner SwettTanner Swett
6,52322 silver badges42 bronze badges
6,52322 silver badges42 bronze badges
$begingroup$
Thank you, this will help, I am wondering if this can be somewhat similarly said for physics or theoretical physics as well
$endgroup$
– C. Jordan
Jul 19 at 13:51
11
$begingroup$
I feel like even though this answer technically provides the direct answer to the OP's question, since the question itself is aimed at better understanding higher dimensional spaces, this answer could be improved by mentioning that while those space do have axes, not everything about axes in 3d applies to higher dimensional spaces. Such as that in dimensions higher than 3, a single axis no longer uniquely determines a rotation.
$endgroup$
– Shufflepants
Jul 19 at 14:59
14
$begingroup$
@Shufflepants - The OP appears to be speaking strictly about coordinate axes. So mixing in the axis of a rotation (which has nothing to do with coordinate axes) is just sowing confusion. And it is only in 3D that you can associate rotations with an axis. In 2D, they are associated with points, not lines, and in 1D rotations don't even exist. More generally, a rotation is associated with an affine space of codimension 2.
$endgroup$
– Paul Sinclair
Jul 19 at 16:58
$begingroup$
What about the vector space $mathbbC^2$ over $mathbbC$?
$endgroup$
– Paracosmiste
Jul 19 at 21:35
add a comment
|
$begingroup$
Thank you, this will help, I am wondering if this can be somewhat similarly said for physics or theoretical physics as well
$endgroup$
– C. Jordan
Jul 19 at 13:51
11
$begingroup$
I feel like even though this answer technically provides the direct answer to the OP's question, since the question itself is aimed at better understanding higher dimensional spaces, this answer could be improved by mentioning that while those space do have axes, not everything about axes in 3d applies to higher dimensional spaces. Such as that in dimensions higher than 3, a single axis no longer uniquely determines a rotation.
$endgroup$
– Shufflepants
Jul 19 at 14:59
14
$begingroup$
@Shufflepants - The OP appears to be speaking strictly about coordinate axes. So mixing in the axis of a rotation (which has nothing to do with coordinate axes) is just sowing confusion. And it is only in 3D that you can associate rotations with an axis. In 2D, they are associated with points, not lines, and in 1D rotations don't even exist. More generally, a rotation is associated with an affine space of codimension 2.
$endgroup$
– Paul Sinclair
Jul 19 at 16:58
$begingroup$
What about the vector space $mathbbC^2$ over $mathbbC$?
$endgroup$
– Paracosmiste
Jul 19 at 21:35
$begingroup$
Thank you, this will help, I am wondering if this can be somewhat similarly said for physics or theoretical physics as well
$endgroup$
– C. Jordan
Jul 19 at 13:51
$begingroup$
Thank you, this will help, I am wondering if this can be somewhat similarly said for physics or theoretical physics as well
$endgroup$
– C. Jordan
Jul 19 at 13:51
11
11
$begingroup$
I feel like even though this answer technically provides the direct answer to the OP's question, since the question itself is aimed at better understanding higher dimensional spaces, this answer could be improved by mentioning that while those space do have axes, not everything about axes in 3d applies to higher dimensional spaces. Such as that in dimensions higher than 3, a single axis no longer uniquely determines a rotation.
$endgroup$
– Shufflepants
Jul 19 at 14:59
$begingroup$
I feel like even though this answer technically provides the direct answer to the OP's question, since the question itself is aimed at better understanding higher dimensional spaces, this answer could be improved by mentioning that while those space do have axes, not everything about axes in 3d applies to higher dimensional spaces. Such as that in dimensions higher than 3, a single axis no longer uniquely determines a rotation.
$endgroup$
– Shufflepants
Jul 19 at 14:59
14
14
$begingroup$
@Shufflepants - The OP appears to be speaking strictly about coordinate axes. So mixing in the axis of a rotation (which has nothing to do with coordinate axes) is just sowing confusion. And it is only in 3D that you can associate rotations with an axis. In 2D, they are associated with points, not lines, and in 1D rotations don't even exist. More generally, a rotation is associated with an affine space of codimension 2.
$endgroup$
– Paul Sinclair
Jul 19 at 16:58
$begingroup$
@Shufflepants - The OP appears to be speaking strictly about coordinate axes. So mixing in the axis of a rotation (which has nothing to do with coordinate axes) is just sowing confusion. And it is only in 3D that you can associate rotations with an axis. In 2D, they are associated with points, not lines, and in 1D rotations don't even exist. More generally, a rotation is associated with an affine space of codimension 2.
$endgroup$
– Paul Sinclair
Jul 19 at 16:58
$begingroup$
What about the vector space $mathbbC^2$ over $mathbbC$?
$endgroup$
– Paracosmiste
Jul 19 at 21:35
$begingroup$
What about the vector space $mathbbC^2$ over $mathbbC$?
$endgroup$
– Paracosmiste
Jul 19 at 21:35
add a comment
|
$begingroup$
Sure. In $n$ dimensions, the $k^textth$ axis, for some $1 leq k leq n$, is the span of the vector whose $k^textth$ component is $1$ and all other components $0$. The problem is that there's not a particularly satisfying geometric visualization of these axes and one largely has to settle for this algebraic abstraction.
answered Jul 19 at 1:51
Kaj HansenKaj Hansen
29k4 gold badges41 silver badges85 bronze badges
$endgroup$
add a comment
|
$begingroup$
Sure. In $n$ dimensions, the $k^textth$ axis, for some $1 leq k leq n$, is the span of the vector whose $k^textth$ component is $1$ and all other components $0$. The problem is that there's not a particularly satisfying geometric visualization of these axes and one largely has to settle for this algebraic abstraction.
answered Jul 19 at 1:51
Kaj HansenKaj Hansen
29k4 gold badges41 silver badges85 bronze badges
$endgroup$
add a comment
|
$begingroup$
Sure. In $n$ dimensions, the $k^textth$ axis, for some $1 leq k leq n$, is the span of the vector whose $k^textth$ component is $1$ and all other components $0$. The problem is that there's not a particularly satisfying geometric visualization of these axes and one largely has to settle for this algebraic abstraction.
answered Jul 19 at 1:51
Kaj HansenKaj Hansen
29k4 gold badges41 silver badges85 bronze badges
$endgroup$
Sure. In $n$ dimensions, the $k^textth$ axis, for some $1 leq k leq n$, is the span of the vector whose $k^textth$ component is $1$ and all other components $0$. The problem is that there's not a particularly satisfying geometric visualization of these axes and one largely has to settle for this algebraic abstraction.
answered Jul 19 at 1:51
Kaj HansenKaj Hansen
29k4 gold badges41 silver badges85 bronze badges
edited Jul 20 at 1:41
answered Jul 19 at 1:51
Kaj HansenKaj Hansen
29k4 gold badges41 silver badges85 bronze badges
answered Jul 19 at 1:51
Kaj HansenKaj Hansen
29k4 gold badges41 silver badges85 bronze badges
answered Jul 19 at 1:51
Kaj HansenKaj Hansen
29k4 gold badges41 silver badges85 bronze badges
29k4 gold badges41 silver badges85 bronze badges
add a comment
|
add a comment
|
$begingroup$
The first important term here is that of a vector space. This is a general term; vector spaces can be far more than simply some $mathbbR^n$. Polynomials, for example, form a vector space of infinite dimension.
Now you need to understand what a dimension is. If you have a vector in three dimensions, you write that vector like $(4,1,7)$. Those coordinates make sense in the context of a base. A base is a set of linearly independent vectors, which has the maximal possible size for that vector space. This size is the dimension. A three dimensional space is defined by having maximally three linear independent vectors in a set.
Edit: On the recommendation of Kaj Hansen, let me at this point make clear that a base of a given vector space has always the same amount of vectors. This is important, because this defines the dimension of the vector space.
The coordinates are now in relation to your base. If your base is $b_1, b_2, b_3$, the vector $(4,1,7)$ is the vector you get by the linear combination $4b_1 + 1b_2 + 7b_3$.
Now realize that this base can change. You can take three linear independent vectors different from your current base and make them the new base, resulting in all vectors getting different coordinates.
To give you an example when this is used, consider a 3D image in computer graphics that depicts one of those wooden puppets that artists use. There is a global coordinate system for the whole image, and the puppet has coordinates within that. But the limps have local coordinate systems which are in relation to the joints of the puppet. Hope this helps you.
Now, for the axes, an axis is simply the line created when you follow the direction of a base vector. Nothing more. They are arbitrary, and they do not even have to be orthogonal to each other. They are not a fundamental principle of a space. Coordinates are just numbers how we model it.
Another point of view, let's say that you are to determine the axes of reality, of the three-dimensional space we live in. In which direction would you point the first axis? Arbitrary. You decide on some direction for the first an then for the others, and then you write things in relation to those (or, better to say, the base vectors, as you also need a length). There is not the one objective way to create your axes.
Edit: Inserted from the comment of Kaj Hansen, note that usually one uses the standard base $e_k^n_k=1$ for $mathbbR^n$, which consists of vectors $k_i$ that are all zero except for the $k^th$ component which is one. With those, a coordinate evaluates to a vector with the same numbers. Now, I brought this here and not above for a reason, and that is because we just talked about the "axes of reality", which do not exist. In the real world, there is no objective coordinate (1, 0, 0). For that reason, even the standard base is not "objective" or anything, but bound to some definitions we made. When we compute things in math, (1, 0, 0) is an objective thing, though, because in math, we live only in definitions and axioms.
With more dimensions, it is not different. You have a base and you work with it, and you can totally change the base.
Things might get difficult when you talk about physical spaces, though. Time behaves different as a dimension, and we are now in a Minkowski space, and things are getting complicated.
If you are interested in examples for vector spaces which aren't R^n, like the polynomial vector spaces or that of other functions, I can explain that too, but as this is long enough already, I'll stop for now.
answered Jul 19 at 10:25
AziuthAziuth
1547 bronze badges
$endgroup$
$begingroup$
That is alright, I think I get the picture now, sounds like dimensions do have at least one axis for each one, even for 4 dimensional and beyond
$endgroup$
– C. Jordan
Jul 19 at 13:53
$begingroup$
I salute you for posting this. The word "dimension" is probably the most misunderstood mathematical term that's entered lay vocabulary, and ultimately something like this is what's needed to explain it properly.
$endgroup$
– Kaj Hansen
Jul 20 at 1:39
$begingroup$
Two recommended additions: explicitly point out the facts that 1) any two bases have the same number of vectors; and 2) we generally, unless otherwise stated, use the standard basis $mathbfe_k_k=1^n$ for $mathbbR^n$, where $mathbfe_k$ is the vector whose $k^textth$ component is $1$ and all others $0$. Hence, writing $langle 2, 5, 3 rangle$ is implicitly saying $2 langle 1, 0, 0 rangle + 5 langle 0, 1, 0 rangle + 3 langle 0, 0, 1 rangle$.
$endgroup$
– Kaj Hansen
Jul 20 at 5:03
$begingroup$
@KajHansen Inserted your recommendations. Continued on your second one with some more "philosophical" aspects of math, thought they belong nicely in a post about the basic idea of a dimension.
$endgroup$
– Aziuth
Jul 20 at 8:37
add a comment
|
$begingroup$
The first important term here is that of a vector space. This is a general term; vector spaces can be far more than simply some $mathbbR^n$. Polynomials, for example, form a vector space of infinite dimension.
Now you need to understand what a dimension is. If you have a vector in three dimensions, you write that vector like $(4,1,7)$. Those coordinates make sense in the context of a base. A base is a set of linearly independent vectors, which has the maximal possible size for that vector space. This size is the dimension. A three dimensional space is defined by having maximally three linear independent vectors in a set.
Edit: On the recommendation of Kaj Hansen, let me at this point make clear that a base of a given vector space has always the same amount of vectors. This is important, because this defines the dimension of the vector space.
The coordinates are now in relation to your base. If your base is $b_1, b_2, b_3$, the vector $(4,1,7)$ is the vector you get by the linear combination $4b_1 + 1b_2 + 7b_3$.
Now realize that this base can change. You can take three linear independent vectors different from your current base and make them the new base, resulting in all vectors getting different coordinates.
To give you an example when this is used, consider a 3D image in computer graphics that depicts one of those wooden puppets that artists use. There is a global coordinate system for the whole image, and the puppet has coordinates within that. But the limps have local coordinate systems which are in relation to the joints of the puppet. Hope this helps you.
Now, for the axes, an axis is simply the line created when you follow the direction of a base vector. Nothing more. They are arbitrary, and they do not even have to be orthogonal to each other. They are not a fundamental principle of a space. Coordinates are just numbers how we model it.
Another point of view, let's say that you are to determine the axes of reality, of the three-dimensional space we live in. In which direction would you point the first axis? Arbitrary. You decide on some direction for the first an then for the others, and then you write things in relation to those (or, better to say, the base vectors, as you also need a length). There is not the one objective way to create your axes.
Edit: Inserted from the comment of Kaj Hansen, note that usually one uses the standard base $e_k^n_k=1$ for $mathbbR^n$, which consists of vectors $k_i$ that are all zero except for the $k^th$ component which is one. With those, a coordinate evaluates to a vector with the same numbers. Now, I brought this here and not above for a reason, and that is because we just talked about the "axes of reality", which do not exist. In the real world, there is no objective coordinate (1, 0, 0). For that reason, even the standard base is not "objective" or anything, but bound to some definitions we made. When we compute things in math, (1, 0, 0) is an objective thing, though, because in math, we live only in definitions and axioms.
With more dimensions, it is not different. You have a base and you work with it, and you can totally change the base.
Things might get difficult when you talk about physical spaces, though. Time behaves different as a dimension, and we are now in a Minkowski space, and things are getting complicated.
If you are interested in examples for vector spaces which aren't R^n, like the polynomial vector spaces or that of other functions, I can explain that too, but as this is long enough already, I'll stop for now.
answered Jul 19 at 10:25
AziuthAziuth
1547 bronze badges
$endgroup$
$begingroup$
That is alright, I think I get the picture now, sounds like dimensions do have at least one axis for each one, even for 4 dimensional and beyond
$endgroup$
– C. Jordan
Jul 19 at 13:53
$begingroup$
I salute you for posting this. The word "dimension" is probably the most misunderstood mathematical term that's entered lay vocabulary, and ultimately something like this is what's needed to explain it properly.
$endgroup$
– Kaj Hansen
Jul 20 at 1:39
$begingroup$
Two recommended additions: explicitly point out the facts that 1) any two bases have the same number of vectors; and 2) we generally, unless otherwise stated, use the standard basis $mathbfe_k_k=1^n$ for $mathbbR^n$, where $mathbfe_k$ is the vector whose $k^textth$ component is $1$ and all others $0$. Hence, writing $langle 2, 5, 3 rangle$ is implicitly saying $2 langle 1, 0, 0 rangle + 5 langle 0, 1, 0 rangle + 3 langle 0, 0, 1 rangle$.
$endgroup$
– Kaj Hansen
Jul 20 at 5:03
$begingroup$
@KajHansen Inserted your recommendations. Continued on your second one with some more "philosophical" aspects of math, thought they belong nicely in a post about the basic idea of a dimension.
$endgroup$
– Aziuth
Jul 20 at 8:37
add a comment
|
$begingroup$
The first important term here is that of a vector space. This is a general term; vector spaces can be far more than simply some $mathbbR^n$. Polynomials, for example, form a vector space of infinite dimension.
Now you need to understand what a dimension is. If you have a vector in three dimensions, you write that vector like $(4,1,7)$. Those coordinates make sense in the context of a base. A base is a set of linearly independent vectors, which has the maximal possible size for that vector space. This size is the dimension. A three dimensional space is defined by having maximally three linear independent vectors in a set.
Edit: On the recommendation of Kaj Hansen, let me at this point make clear that a base of a given vector space has always the same amount of vectors. This is important, because this defines the dimension of the vector space.
The coordinates are now in relation to your base. If your base is $b_1, b_2, b_3$, the vector $(4,1,7)$ is the vector you get by the linear combination $4b_1 + 1b_2 + 7b_3$.
Now realize that this base can change. You can take three linear independent vectors different from your current base and make them the new base, resulting in all vectors getting different coordinates.
To give you an example when this is used, consider a 3D image in computer graphics that depicts one of those wooden puppets that artists use. There is a global coordinate system for the whole image, and the puppet has coordinates within that. But the limps have local coordinate systems which are in relation to the joints of the puppet. Hope this helps you.
Now, for the axes, an axis is simply the line created when you follow the direction of a base vector. Nothing more. They are arbitrary, and they do not even have to be orthogonal to each other. They are not a fundamental principle of a space. Coordinates are just numbers how we model it.
Another point of view, let's say that you are to determine the axes of reality, of the three-dimensional space we live in. In which direction would you point the first axis? Arbitrary. You decide on some direction for the first an then for the others, and then you write things in relation to those (or, better to say, the base vectors, as you also need a length). There is not the one objective way to create your axes.
Edit: Inserted from the comment of Kaj Hansen, note that usually one uses the standard base $e_k^n_k=1$ for $mathbbR^n$, which consists of vectors $k_i$ that are all zero except for the $k^th$ component which is one. With those, a coordinate evaluates to a vector with the same numbers. Now, I brought this here and not above for a reason, and that is because we just talked about the "axes of reality", which do not exist. In the real world, there is no objective coordinate (1, 0, 0). For that reason, even the standard base is not "objective" or anything, but bound to some definitions we made. When we compute things in math, (1, 0, 0) is an objective thing, though, because in math, we live only in definitions and axioms.
With more dimensions, it is not different. You have a base and you work with it, and you can totally change the base.
Things might get difficult when you talk about physical spaces, though. Time behaves different as a dimension, and we are now in a Minkowski space, and things are getting complicated.
If you are interested in examples for vector spaces which aren't R^n, like the polynomial vector spaces or that of other functions, I can explain that too, but as this is long enough already, I'll stop for now.
answered Jul 19 at 10:25
AziuthAziuth
1547 bronze badges
$endgroup$
The first important term here is that of a vector space. This is a general term; vector spaces can be far more than simply some $mathbbR^n$. Polynomials, for example, form a vector space of infinite dimension.
Now you need to understand what a dimension is. If you have a vector in three dimensions, you write that vector like $(4,1,7)$. Those coordinates make sense in the context of a base. A base is a set of linearly independent vectors, which has the maximal possible size for that vector space. This size is the dimension. A three dimensional space is defined by having maximally three linear independent vectors in a set.
Edit: On the recommendation of Kaj Hansen, let me at this point make clear that a base of a given vector space has always the same amount of vectors. This is important, because this defines the dimension of the vector space.
The coordinates are now in relation to your base. If your base is $b_1, b_2, b_3$, the vector $(4,1,7)$ is the vector you get by the linear combination $4b_1 + 1b_2 + 7b_3$.
Now realize that this base can change. You can take three linear independent vectors different from your current base and make them the new base, resulting in all vectors getting different coordinates.
To give you an example when this is used, consider a 3D image in computer graphics that depicts one of those wooden puppets that artists use. There is a global coordinate system for the whole image, and the puppet has coordinates within that. But the limps have local coordinate systems which are in relation to the joints of the puppet. Hope this helps you.
Now, for the axes, an axis is simply the line created when you follow the direction of a base vector. Nothing more. They are arbitrary, and they do not even have to be orthogonal to each other. They are not a fundamental principle of a space. Coordinates are just numbers how we model it.
Another point of view, let's say that you are to determine the axes of reality, of the three-dimensional space we live in. In which direction would you point the first axis? Arbitrary. You decide on some direction for the first an then for the others, and then you write things in relation to those (or, better to say, the base vectors, as you also need a length). There is not the one objective way to create your axes.
Edit: Inserted from the comment of Kaj Hansen, note that usually one uses the standard base $e_k^n_k=1$ for $mathbbR^n$, which consists of vectors $k_i$ that are all zero except for the $k^th$ component which is one. With those, a coordinate evaluates to a vector with the same numbers. Now, I brought this here and not above for a reason, and that is because we just talked about the "axes of reality", which do not exist. In the real world, there is no objective coordinate (1, 0, 0). For that reason, even the standard base is not "objective" or anything, but bound to some definitions we made. When we compute things in math, (1, 0, 0) is an objective thing, though, because in math, we live only in definitions and axioms.
With more dimensions, it is not different. You have a base and you work with it, and you can totally change the base.
Things might get difficult when you talk about physical spaces, though. Time behaves different as a dimension, and we are now in a Minkowski space, and things are getting complicated.
If you are interested in examples for vector spaces which aren't R^n, like the polynomial vector spaces or that of other functions, I can explain that too, but as this is long enough already, I'll stop for now.
answered Jul 19 at 10:25
AziuthAziuth
1547 bronze badges
edited Jul 20 at 8:42
answered Jul 19 at 10:25
AziuthAziuth
1547 bronze badges
answered Jul 19 at 10:25
AziuthAziuth
1547 bronze badges
answered Jul 19 at 10:25
AziuthAziuth
1547 bronze badges
1547 bronze badges
$begingroup$
That is alright, I think I get the picture now, sounds like dimensions do have at least one axis for each one, even for 4 dimensional and beyond
$endgroup$
– C. Jordan
Jul 19 at 13:53
$begingroup$
I salute you for posting this. The word "dimension" is probably the most misunderstood mathematical term that's entered lay vocabulary, and ultimately something like this is what's needed to explain it properly.
$endgroup$
– Kaj Hansen
Jul 20 at 1:39
$begingroup$
Two recommended additions: explicitly point out the facts that 1) any two bases have the same number of vectors; and 2) we generally, unless otherwise stated, use the standard basis $mathbfe_k_k=1^n$ for $mathbbR^n$, where $mathbfe_k$ is the vector whose $k^textth$ component is $1$ and all others $0$. Hence, writing $langle 2, 5, 3 rangle$ is implicitly saying $2 langle 1, 0, 0 rangle + 5 langle 0, 1, 0 rangle + 3 langle 0, 0, 1 rangle$.
$endgroup$
– Kaj Hansen
Jul 20 at 5:03
$begingroup$
@KajHansen Inserted your recommendations. Continued on your second one with some more "philosophical" aspects of math, thought they belong nicely in a post about the basic idea of a dimension.
$endgroup$
– Aziuth
Jul 20 at 8:37
add a comment
|
$begingroup$
That is alright, I think I get the picture now, sounds like dimensions do have at least one axis for each one, even for 4 dimensional and beyond
$endgroup$
– C. Jordan
Jul 19 at 13:53
$begingroup$
I salute you for posting this. The word "dimension" is probably the most misunderstood mathematical term that's entered lay vocabulary, and ultimately something like this is what's needed to explain it properly.
$endgroup$
– Kaj Hansen
Jul 20 at 1:39
$begingroup$
Two recommended additions: explicitly point out the facts that 1) any two bases have the same number of vectors; and 2) we generally, unless otherwise stated, use the standard basis $mathbfe_k_k=1^n$ for $mathbbR^n$, where $mathbfe_k$ is the vector whose $k^textth$ component is $1$ and all others $0$. Hence, writing $langle 2, 5, 3 rangle$ is implicitly saying $2 langle 1, 0, 0 rangle + 5 langle 0, 1, 0 rangle + 3 langle 0, 0, 1 rangle$.
$endgroup$
– Kaj Hansen
Jul 20 at 5:03
$begingroup$
@KajHansen Inserted your recommendations. Continued on your second one with some more "philosophical" aspects of math, thought they belong nicely in a post about the basic idea of a dimension.
$endgroup$
– Aziuth
Jul 20 at 8:37
$begingroup$
That is alright, I think I get the picture now, sounds like dimensions do have at least one axis for each one, even for 4 dimensional and beyond
$endgroup$
– C. Jordan
Jul 19 at 13:53
$begingroup$
That is alright, I think I get the picture now, sounds like dimensions do have at least one axis for each one, even for 4 dimensional and beyond
$endgroup$
– C. Jordan
Jul 19 at 13:53
$begingroup$
I salute you for posting this. The word "dimension" is probably the most misunderstood mathematical term that's entered lay vocabulary, and ultimately something like this is what's needed to explain it properly.
$endgroup$
– Kaj Hansen
Jul 20 at 1:39
$begingroup$
I salute you for posting this. The word "dimension" is probably the most misunderstood mathematical term that's entered lay vocabulary, and ultimately something like this is what's needed to explain it properly.
$endgroup$
– Kaj Hansen
Jul 20 at 1:39
$begingroup$
Two recommended additions: explicitly point out the facts that 1) any two bases have the same number of vectors; and 2) we generally, unless otherwise stated, use the standard basis $mathbfe_k_k=1^n$ for $mathbbR^n$, where $mathbfe_k$ is the vector whose $k^textth$ component is $1$ and all others $0$. Hence, writing $langle 2, 5, 3 rangle$ is implicitly saying $2 langle 1, 0, 0 rangle + 5 langle 0, 1, 0 rangle + 3 langle 0, 0, 1 rangle$.
$endgroup$
– Kaj Hansen
Jul 20 at 5:03
$begingroup$
Two recommended additions: explicitly point out the facts that 1) any two bases have the same number of vectors; and 2) we generally, unless otherwise stated, use the standard basis $mathbfe_k_k=1^n$ for $mathbbR^n$, where $mathbfe_k$ is the vector whose $k^textth$ component is $1$ and all others $0$. Hence, writing $langle 2, 5, 3 rangle$ is implicitly saying $2 langle 1, 0, 0 rangle + 5 langle 0, 1, 0 rangle + 3 langle 0, 0, 1 rangle$.
$endgroup$
– Kaj Hansen
Jul 20 at 5:03
$begingroup$
@KajHansen Inserted your recommendations. Continued on your second one with some more "philosophical" aspects of math, thought they belong nicely in a post about the basic idea of a dimension.
$endgroup$
– Aziuth
Jul 20 at 8:37
$begingroup$
@KajHansen Inserted your recommendations. Continued on your second one with some more "philosophical" aspects of math, thought they belong nicely in a post about the basic idea of a dimension.
$endgroup$
– Aziuth
Jul 20 at 8:37
add a comment
|
$begingroup$
In three dimensions the three axes each perpendicular to the other two tell you that you identify a point by giving its coordinates $(x,y,z)$.
The three axes are the lists of coordinates that have just one entry that's not $0$.
Then you can do geometry by doing algebra with the coordinates.
For example, the eight points
(0, 0, 0) (1, 1, 0)
(1, 0, 0) (1, 0, 1)
(0, 1, 0) (0, 1, 1)
(0, 0, 1) (1, 1, 1)
are the vertices of a cube. (Those are all the lists of coordinates you can make using just $0$ and $1$.)
If you just allow yourself more coordinates you can study higher dimensions. So the $16$ ways to make all the lists of four coordinates each of which is just $0$ and $1$ are the vertices of the analogue of a cube in four dimensions. You can study its properties. Actually "seeing" it as a geometric object is pretty much beyond our human abilities. Here's the best we can do: https://en.wikipedia.org/wiki/Tesseract .
Clearly there's no need to stop at four coordinates/dimensions.
answered Jul 19 at 2:04
Ethan BolkerEthan Bolker
58.6k5 gold badges64 silver badges140 bronze badges
$endgroup$
add a comment
|
$begingroup$
In three dimensions the three axes each perpendicular to the other two tell you that you identify a point by giving its coordinates $(x,y,z)$.
The three axes are the lists of coordinates that have just one entry that's not $0$.
Then you can do geometry by doing algebra with the coordinates.
For example, the eight points
(0, 0, 0) (1, 1, 0)
(1, 0, 0) (1, 0, 1)
(0, 1, 0) (0, 1, 1)
(0, 0, 1) (1, 1, 1)
are the vertices of a cube. (Those are all the lists of coordinates you can make using just $0$ and $1$.)
If you just allow yourself more coordinates you can study higher dimensions. So the $16$ ways to make all the lists of four coordinates each of which is just $0$ and $1$ are the vertices of the analogue of a cube in four dimensions. You can study its properties. Actually "seeing" it as a geometric object is pretty much beyond our human abilities. Here's the best we can do: https://en.wikipedia.org/wiki/Tesseract .
Clearly there's no need to stop at four coordinates/dimensions.
answered Jul 19 at 2:04
Ethan BolkerEthan Bolker
58.6k5 gold badges64 silver badges140 bronze badges
$endgroup$
add a comment
|
$begingroup$
In three dimensions the three axes each perpendicular to the other two tell you that you identify a point by giving its coordinates $(x,y,z)$.
The three axes are the lists of coordinates that have just one entry that's not $0$.
Then you can do geometry by doing algebra with the coordinates.
For example, the eight points
(0, 0, 0) (1, 1, 0)
(1, 0, 0) (1, 0, 1)
(0, 1, 0) (0, 1, 1)
(0, 0, 1) (1, 1, 1)
are the vertices of a cube. (Those are all the lists of coordinates you can make using just $0$ and $1$.)
If you just allow yourself more coordinates you can study higher dimensions. So the $16$ ways to make all the lists of four coordinates each of which is just $0$ and $1$ are the vertices of the analogue of a cube in four dimensions. You can study its properties. Actually "seeing" it as a geometric object is pretty much beyond our human abilities. Here's the best we can do: https://en.wikipedia.org/wiki/Tesseract .
Clearly there's no need to stop at four coordinates/dimensions.
answered Jul 19 at 2:04
Ethan BolkerEthan Bolker
58.6k5 gold badges64 silver badges140 bronze badges
$endgroup$
In three dimensions the three axes each perpendicular to the other two tell you that you identify a point by giving its coordinates $(x,y,z)$.
The three axes are the lists of coordinates that have just one entry that's not $0$.
Then you can do geometry by doing algebra with the coordinates.
For example, the eight points
(0, 0, 0) (1, 1, 0)
(1, 0, 0) (1, 0, 1)
(0, 1, 0) (0, 1, 1)
(0, 0, 1) (1, 1, 1)
are the vertices of a cube. (Those are all the lists of coordinates you can make using just $0$ and $1$.)
If you just allow yourself more coordinates you can study higher dimensions. So the $16$ ways to make all the lists of four coordinates each of which is just $0$ and $1$ are the vertices of the analogue of a cube in four dimensions. You can study its properties. Actually "seeing" it as a geometric object is pretty much beyond our human abilities. Here's the best we can do: https://en.wikipedia.org/wiki/Tesseract .
Clearly there's no need to stop at four coordinates/dimensions.
answered Jul 19 at 2:04
Ethan BolkerEthan Bolker
58.6k5 gold badges64 silver badges140 bronze badges
answered Jul 19 at 2:04
Ethan BolkerEthan Bolker
58.6k5 gold badges64 silver badges140 bronze badges
answered Jul 19 at 2:04
Ethan BolkerEthan Bolker
58.6k5 gold badges64 silver badges140 bronze badges
answered Jul 19 at 2:04
Ethan BolkerEthan Bolker
58.6k5 gold badges64 silver badges140 bronze badges
58.6k5 gold badges64 silver badges140 bronze badges
add a comment
|
add a comment
|
$begingroup$
Yes. Forget everything you previously saw with the cartesian coordinate system, etc.
Now, imagine a ball of yarn with a bunch of needles sticking out of it. Those dimensions each have a needle (an axis for that dimension). Now, when you have a bunch of dimensions, each thing you add to your system needs to have a thread of yarn that touches each needle somewhere along its length.
Each thread is a vector, or the position of that thing you added to your system.
answered Jul 19 at 20:04
JakeJJakeJ
1291 bronze badge
$endgroup$
add a comment
|
$begingroup$
Yes. Forget everything you previously saw with the cartesian coordinate system, etc.
Now, imagine a ball of yarn with a bunch of needles sticking out of it. Those dimensions each have a needle (an axis for that dimension). Now, when you have a bunch of dimensions, each thing you add to your system needs to have a thread of yarn that touches each needle somewhere along its length.
Each thread is a vector, or the position of that thing you added to your system.
answered Jul 19 at 20:04
JakeJJakeJ
1291 bronze badge
$endgroup$
add a comment
|
$begingroup$
Yes. Forget everything you previously saw with the cartesian coordinate system, etc.
Now, imagine a ball of yarn with a bunch of needles sticking out of it. Those dimensions each have a needle (an axis for that dimension). Now, when you have a bunch of dimensions, each thing you add to your system needs to have a thread of yarn that touches each needle somewhere along its length.
Each thread is a vector, or the position of that thing you added to your system.
answered Jul 19 at 20:04
JakeJJakeJ
1291 bronze badge
$endgroup$
Yes. Forget everything you previously saw with the cartesian coordinate system, etc.
Now, imagine a ball of yarn with a bunch of needles sticking out of it. Those dimensions each have a needle (an axis for that dimension). Now, when you have a bunch of dimensions, each thing you add to your system needs to have a thread of yarn that touches each needle somewhere along its length.
Each thread is a vector, or the position of that thing you added to your system.
answered Jul 19 at 20:04
JakeJJakeJ
1291 bronze badge
answered Jul 19 at 20:04
JakeJJakeJ
1291 bronze badge
answered Jul 19 at 20:04
JakeJJakeJ
1291 bronze badge
answered Jul 19 at 20:04
JakeJJakeJ
1291 bronze badge
1291 bronze badge
add a comment
|
add a comment
|
$begingroup$
Each dimension will, by definition, have an axis that is 90 degrees askew to all other dimensions. Since our everyday world is 3-dimensional, this may be difficult to envision/imagine for axes beyond the basic x-y-z, but mathematically this is how it works.
answered Jul 19 at 17:10
EJSawyerEJSawyer
191 bronze badge
$endgroup$
$begingroup$
Use MathJax please
$endgroup$
– Aqua
Jul 19 at 17:39
$begingroup$
Thank you very much, this is what I wish to hear
$endgroup$
– C. Jordan
Jul 19 at 18:27
$begingroup$
From where do the 90 degrees come from? Any subspace is a proper space, no matter how it is angled to some axis. Plus, you make it sound as if a "dimension" is something particular, rather than a number of independencies.
$endgroup$
– Aziuth
Jul 20 at 8:41
add a comment
|
$begingroup$
Each dimension will, by definition, have an axis that is 90 degrees askew to all other dimensions. Since our everyday world is 3-dimensional, this may be difficult to envision/imagine for axes beyond the basic x-y-z, but mathematically this is how it works.
answered Jul 19 at 17:10
EJSawyerEJSawyer
191 bronze badge
$endgroup$
$begingroup$
Use MathJax please
$endgroup$
– Aqua
Jul 19 at 17:39
$begingroup$
Thank you very much, this is what I wish to hear
$endgroup$
– C. Jordan
Jul 19 at 18:27
$begingroup$
From where do the 90 degrees come from? Any subspace is a proper space, no matter how it is angled to some axis. Plus, you make it sound as if a "dimension" is something particular, rather than a number of independencies.
$endgroup$
– Aziuth
Jul 20 at 8:41
add a comment
|
$begingroup$
Each dimension will, by definition, have an axis that is 90 degrees askew to all other dimensions. Since our everyday world is 3-dimensional, this may be difficult to envision/imagine for axes beyond the basic x-y-z, but mathematically this is how it works.
answered Jul 19 at 17:10
EJSawyerEJSawyer
191 bronze badge
$endgroup$
Each dimension will, by definition, have an axis that is 90 degrees askew to all other dimensions. Since our everyday world is 3-dimensional, this may be difficult to envision/imagine for axes beyond the basic x-y-z, but mathematically this is how it works.
answered Jul 19 at 17:10
EJSawyerEJSawyer
191 bronze badge
answered Jul 19 at 17:10
EJSawyerEJSawyer
191 bronze badge
answered Jul 19 at 17:10
EJSawyerEJSawyer
191 bronze badge
answered Jul 19 at 17:10
EJSawyerEJSawyer
191 bronze badge
191 bronze badge
$begingroup$
Use MathJax please
$endgroup$
– Aqua
Jul 19 at 17:39
$begingroup$
Thank you very much, this is what I wish to hear
$endgroup$
– C. Jordan
Jul 19 at 18:27
$begingroup$
From where do the 90 degrees come from? Any subspace is a proper space, no matter how it is angled to some axis. Plus, you make it sound as if a "dimension" is something particular, rather than a number of independencies.
$endgroup$
– Aziuth
Jul 20 at 8:41
add a comment
|
$begingroup$
Use MathJax please
$endgroup$
– Aqua
Jul 19 at 17:39
$begingroup$
Thank you very much, this is what I wish to hear
$endgroup$
– C. Jordan
Jul 19 at 18:27
$begingroup$
From where do the 90 degrees come from? Any subspace is a proper space, no matter how it is angled to some axis. Plus, you make it sound as if a "dimension" is something particular, rather than a number of independencies.
$endgroup$
– Aziuth
Jul 20 at 8:41
$begingroup$
Use MathJax please
$endgroup$
– Aqua
Jul 19 at 17:39
$begingroup$
Use MathJax please
$endgroup$
– Aqua
Jul 19 at 17:39
$begingroup$
Thank you very much, this is what I wish to hear
$endgroup$
– C. Jordan
Jul 19 at 18:27
$begingroup$
Thank you very much, this is what I wish to hear
$endgroup$
– C. Jordan
Jul 19 at 18:27
$begingroup$
From where do the 90 degrees come from? Any subspace is a proper space, no matter how it is angled to some axis. Plus, you make it sound as if a "dimension" is something particular, rather than a number of independencies.
$endgroup$
– Aziuth
Jul 20 at 8:41
$begingroup$
From where do the 90 degrees come from? Any subspace is a proper space, no matter how it is angled to some axis. Plus, you make it sound as if a "dimension" is something particular, rather than a number of independencies.
$endgroup$
– Aziuth
Jul 20 at 8:41
add a comment
|
$begingroup$
Proof: Suppose $n$ is the highest dimension for which we have axes. Let $(x_1,x_2,ldots, x_n)$ be any arbitrary point in this dimension. By assumption there are no axes in $n+1$ dimension, so there must be infinitely many points with the exact same coordinates $(x_1,x_2,ldots, x_n,666)$ i.e. unique points in $n+1$ dimension do not exist which is a contradiction to the definition of a dimension.
answered Jul 20 at 1:53
Nilotpal Kanti SinhaNilotpal Kanti Sinha
7,7802 gold badges19 silver badges52 bronze badges
$endgroup$
add a comment
|
$begingroup$
Proof: Suppose $n$ is the highest dimension for which we have axes. Let $(x_1,x_2,ldots, x_n)$ be any arbitrary point in this dimension. By assumption there are no axes in $n+1$ dimension, so there must be infinitely many points with the exact same coordinates $(x_1,x_2,ldots, x_n,666)$ i.e. unique points in $n+1$ dimension do not exist which is a contradiction to the definition of a dimension.
answered Jul 20 at 1:53
Nilotpal Kanti SinhaNilotpal Kanti Sinha
7,7802 gold badges19 silver badges52 bronze badges
$endgroup$
add a comment
|
$begingroup$
Proof: Suppose $n$ is the highest dimension for which we have axes. Let $(x_1,x_2,ldots, x_n)$ be any arbitrary point in this dimension. By assumption there are no axes in $n+1$ dimension, so there must be infinitely many points with the exact same coordinates $(x_1,x_2,ldots, x_n,666)$ i.e. unique points in $n+1$ dimension do not exist which is a contradiction to the definition of a dimension.
answered Jul 20 at 1:53
Nilotpal Kanti SinhaNilotpal Kanti Sinha
7,7802 gold badges19 silver badges52 bronze badges
$endgroup$
Proof: Suppose $n$ is the highest dimension for which we have axes. Let $(x_1,x_2,ldots, x_n)$ be any arbitrary point in this dimension. By assumption there are no axes in $n+1$ dimension, so there must be infinitely many points with the exact same coordinates $(x_1,x_2,ldots, x_n,666)$ i.e. unique points in $n+1$ dimension do not exist which is a contradiction to the definition of a dimension.
answered Jul 20 at 1:53
Nilotpal Kanti SinhaNilotpal Kanti Sinha
7,7802 gold badges19 silver badges52 bronze badges
edited Jul 20 at 1:58
answered Jul 20 at 1:53
Nilotpal Kanti SinhaNilotpal Kanti Sinha
7,7802 gold badges19 silver badges52 bronze badges
answered Jul 20 at 1:53
Nilotpal Kanti SinhaNilotpal Kanti Sinha
7,7802 gold badges19 silver badges52 bronze badges
answered Jul 20 at 1:53
Nilotpal Kanti SinhaNilotpal Kanti Sinha
7,7802 gold badges19 silver badges52 bronze badges
7,7802 gold badges19 silver badges52 bronze badges
add a comment
|
add a comment
|
3
$begingroup$
what do you mean by an axis for volume?
$endgroup$
– Mirko
Jul 19 at 3:14
$begingroup$
@Mirko as in three dimensional space
$endgroup$
– C. Jordan
Jul 19 at 3:36
15
$begingroup$
Heh heh. I am thinking about a higher-dimensional hero swinging an ax at the head of a Killing Form to defeat the Lie Algebra.
$endgroup$
– puppetsock
Jul 19 at 13:42
4
$begingroup$
I'd say that's pretty much the definition of a dimension - think of it as another set of directions you can go. 1D lets you go only left and right, 2D adds forward and backward, 3D adds up and down, etc. Each dimension is just like the previous, only with another axis at a right angle to all the previous axes. (Obviously this becomes difficult to visualize past 3, but the math doesn't really care about that, it works pretty much the same for any number of dimensions.)
$endgroup$
– Darrel Hoffman
Jul 19 at 14:29
$begingroup$
@DarrelHoffman I don't see what makes it so hard to visualize. I just use a trick that my high school pre-calculus teacher taught me ages ago: if you want to visualize a four dimensional space, first just visualize an $n$-dimensional space, then take $n=4$. What could be simpler? :P
$endgroup$
– Xander Henderson
Jul 20 at 13:09