Why aren't tangent spaces simply defined as vector spaces with same dimension as the manifold?Prove that the dimension of the tangent space $T_x(X)$ of a k-dimensional manifold is kQuestion about Milnor's talk at the Abel PrizeWhat exactly is a Kähler Manifold?Prove that the tangent space has the same dimension as the manifoldDifferential topology versus differential geometryTangent spaces and derivatives are defined in the setting of manifolds with boundaryTopological Manifolds: Why parametrization is essential?tangent bundle projection is $C^infty$ even though the manifold is only $C^k+1$?
Was Haile Selassie the world's only involuntary messiah?
Ross 128 (What does the number mean?)
Beautiful planar geometry theorems not encountered in high school
SSD not reaching advertised speed
How to calculate number of sets in Sigma Algebra
What does play with feeling mean?
At what point should I switch between a resistor divider and a LDO?
Locked folder with obscure app from SourceForge - now I cannot unlock the folder
Hand propping Cessna 172N, with a pilot and a non intuned passenger
How can I find out where to buy uncommon (for the location) items while traveling?
SSH host identification changes on one wireless network
Seen from Europe, why is there a hard separation between Republicans and Democrats in the US?
Is "montäglich" commonly used?
Hot Rim Looking for Suggestions
Taxes on rent back
My PhD defense is next week and I am having negative thoughts about my work and knowledge. Any advice on how to tackle this?
Why is there potato in meatballs?
How does the credit rating of a country affect its populace?
How does mounted combatant interact with total cover?
Is it acceptable the secretariat to have full access to our entire Outlook agenda?
CEO says not to expect pay increases unless you do something really exceptional. Is this counter-productive?
Is it a mistake to use a password that has previously been used (by anyone ever)?
If a photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?
Car imitates dead battery but comes back to life ~30 minutes later and lets me start it
Why aren't tangent spaces simply defined as vector spaces with same dimension as the manifold?
Prove that the dimension of the tangent space $T_x(X)$ of a k-dimensional manifold is kQuestion about Milnor's talk at the Abel PrizeWhat exactly is a Kähler Manifold?Prove that the tangent space has the same dimension as the manifoldDifferential topology versus differential geometryTangent spaces and derivatives are defined in the setting of manifolds with boundaryTopological Manifolds: Why parametrization is essential?tangent bundle projection is $C^infty$ even though the manifold is only $C^k+1$?
.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'm a physics student starting grad school, and I figured I'd read up on manifolds since they pop up so much. However, one thing that continues to elude me is why tangent spaces have such involved definitions. Given that the tangent space of an $n$ dimensional manifold at any point is diffeomorphic to $mathbbR^n$, why do we bother with all the long definitions involving derivations and equivalence classes of curves in the first place?
differential-topology tangent-spaces
asked Sep 11 at 2:36
el duderinoel duderino
2451 silver badge3 bronze badges
$endgroup$
add a comment
|
$begingroup$
I'm a physics student starting grad school, and I figured I'd read up on manifolds since they pop up so much. However, one thing that continues to elude me is why tangent spaces have such involved definitions. Given that the tangent space of an $n$ dimensional manifold at any point is diffeomorphic to $mathbbR^n$, why do we bother with all the long definitions involving derivations and equivalence classes of curves in the first place?
differential-topology tangent-spaces
asked Sep 11 at 2:36
el duderinoel duderino
2451 silver badge3 bronze badges
$endgroup$
2
$begingroup$
The tangent space of a smooth, $n$-dimensional manifold is diffeomorphic to $R^n$, but not canonically: to produce such a diffeomorphism, you need to choose coordinates. The utility of the sophisticated definition of the tangent space comes from the fact that it is intrinsic, i.e., does not depend on any auxiliary choices.
$endgroup$
– fmg
Sep 11 at 3:59
add a comment
|
$begingroup$
I'm a physics student starting grad school, and I figured I'd read up on manifolds since they pop up so much. However, one thing that continues to elude me is why tangent spaces have such involved definitions. Given that the tangent space of an $n$ dimensional manifold at any point is diffeomorphic to $mathbbR^n$, why do we bother with all the long definitions involving derivations and equivalence classes of curves in the first place?
differential-topology tangent-spaces
asked Sep 11 at 2:36
el duderinoel duderino
2451 silver badge3 bronze badges
$endgroup$
I'm a physics student starting grad school, and I figured I'd read up on manifolds since they pop up so much. However, one thing that continues to elude me is why tangent spaces have such involved definitions. Given that the tangent space of an $n$ dimensional manifold at any point is diffeomorphic to $mathbbR^n$, why do we bother with all the long definitions involving derivations and equivalence classes of curves in the first place?
differential-topology tangent-spaces
differential-topology tangent-spaces
asked Sep 11 at 2:36
el duderinoel duderino
2451 silver badge3 bronze badges
asked Sep 11 at 2:36
el duderinoel duderino
2451 silver badge3 bronze badges
asked Sep 11 at 2:36
el duderinoel duderino
2451 silver badge3 bronze badges
asked Sep 11 at 2:36
el duderinoel duderino
2451 silver badge3 bronze badges
asked Sep 11 at 2:36
el duderinoel duderino
2451 silver badge3 bronze badges
2451 silver badge3 bronze badges
2
$begingroup$
The tangent space of a smooth, $n$-dimensional manifold is diffeomorphic to $R^n$, but not canonically: to produce such a diffeomorphism, you need to choose coordinates. The utility of the sophisticated definition of the tangent space comes from the fact that it is intrinsic, i.e., does not depend on any auxiliary choices.
$endgroup$
– fmg
Sep 11 at 3:59
add a comment
|
2
$begingroup$
The tangent space of a smooth, $n$-dimensional manifold is diffeomorphic to $R^n$, but not canonically: to produce such a diffeomorphism, you need to choose coordinates. The utility of the sophisticated definition of the tangent space comes from the fact that it is intrinsic, i.e., does not depend on any auxiliary choices.
$endgroup$
– fmg
Sep 11 at 3:59
2
2
$begingroup$
The tangent space of a smooth, $n$-dimensional manifold is diffeomorphic to $R^n$, but not canonically: to produce such a diffeomorphism, you need to choose coordinates. The utility of the sophisticated definition of the tangent space comes from the fact that it is intrinsic, i.e., does not depend on any auxiliary choices.
$endgroup$
– fmg
Sep 11 at 3:59
$begingroup$
The tangent space of a smooth, $n$-dimensional manifold is diffeomorphic to $R^n$, but not canonically: to produce such a diffeomorphism, you need to choose coordinates. The utility of the sophisticated definition of the tangent space comes from the fact that it is intrinsic, i.e., does not depend on any auxiliary choices.
$endgroup$
– fmg
Sep 11 at 3:59
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
We don't just want to have a vector space to call the "tangent space". We want to do geometric things with the tangent space, and we can't do those things if it's just an arbitrary vector space of the right dimension. We need it to more specifically be a vector space that encodes the parts of the geometry of our manifold that we care about.
For instance, here's one geometric thing we'd like to be able to do: given a smooth curve $gamma:mathbbRto M$ on a manifold $M$, we'd like to define a "tangent vector" at each point of the curve. For each $tinmathbbR$, this should give us a vector "$gamma'(t)$" which is in the tangent space of $M$ at the point $gamma(t)$. If the tangent space is just defined as some arbitrary $n$-dimensional vector space, there isn't going to be any natural way to define $gamma'(t)$. But for the usual definition, there is: we just take the equivalence class of the curve $gamma$.
Another very useful thing we like to do is differentiate functions between manifolds: if $f:Mto N$ is a smooth function between manifolds and $pin M$, there should be a "derivative" $df$ which is a linear map from $T_pMto T_f(p)N$. Again, there's not any natural way to get such a map if the tangent spaces are just some arbitrary vector spaces.
To put it another way, the tangent space to a manifold at a point is not merely a vector space. It is a vector space together with a bunch of extra structure relating it to other geometric features of the manifold. So we care about it not just as a vector space up to isomorphism, but as a "vector space plus extra structure" up to isomorphism. It's possible to axiomatize this extra structure, and then you could define the tangent space as just any abstract gadget satisfying those axioms, rather than restricting yourself to the usual definition by equivalence classes of curves. This isn't commonly done though because (as far as I know) there isn't any nice axiomatization that isn't essentially just saying "a vector space equipped with an isomorphism to this particular concrete construction", so you might as well just use the concrete construction itself.
answered Sep 11 at 2:49
Eric WofseyEric Wofsey
220k15 gold badges256 silver badges399 bronze badges
$endgroup$
$begingroup$
Do you have a reference for axiomatizing this “extra structure” on tangent spaces?
$endgroup$
– Santana Afton
Sep 11 at 4:20
$begingroup$
@SantanaAfton: No, I don't recall ever seeing such a thing written down. Of course, one way to specify the "extra structure" is to just say you have a chosen isomorphism to your favorite specific construction of the tangent space. Any definition is going to be essentially equivalent to this (since of course you want the thing you're defining to be canonically isomorphic to the specific construction), and I don't know of a definition that is really different on its face.
$endgroup$
– Eric Wofsey
Sep 11 at 4:24
$begingroup$
Here's an analogy: you can construct real numbers as Dedekind cuts of rationals. You could also axiomatize real numbers as an ordered field which has exactly one element in each Dedekind cut of the copy of the rationals sitting inside it. But this isn't a particularly useful axiomatization, since it's basically just saying you have an ordered field isomorphic to the ordered field Dedekind cuts of the the rationals. A more useful axiomatization would be that the reals are a complete ordered field.
$endgroup$
– Eric Wofsey
Sep 11 at 4:29
$begingroup$
So, in the case of tangent spaces, it's easy to write down not-useful axiomatizations analogous to that first axiomatization of real numbers, but I don't know of a nice axiomatization analogous the second one. (And if such a thing existed, I suspect it would be used fairly often!)
$endgroup$
– Eric Wofsey
Sep 11 at 4:30
2
$begingroup$
@EricWofsey The definition of the tangent space as equivalence classes of derivations comes to mind, i.e. of operators $D$ on the continuous functions on your manifold that satisfy $D(fg) = fD(g) + D(f)g$. There is no mention of curves or vectors in there so you might be able to spin some substantially different axiomatization from there.
$endgroup$
– mlk
Sep 11 at 12:27
|
show 2 more comments
$begingroup$
If they are “just vector spaces” at each point, then every set would be a manifold.
The whole point is to link the topological structure of the manifold (using coordinate patches that fit together well) to linear algebra that goes on in the tangent spaces.
Another reason to give many different definitions is that they all expose some aspect of what a manifold does. In particular, I’ve found D. Holm’s chapter on manifolds in Geometric mechanics and symmetry very helpful, as it relates several equivalent definitions.
Finally, another contribution to your angst might be the books you’re reading. Some math expositions by physicists can be extremely frustrating for mathematicians, due to different styles in the two disciplines (to put it charitably), so if you’re reading books by such authors it may contribute to what you’re describing.
answered Sep 11 at 2:42
rschwiebrschwieb
116k13 gold badges114 silver badges269 bronze badges
$endgroup$
$begingroup$
+1 for "different styles" point. Very true. Matthew Mathematician says "First assume the following axioms ..."; Peter Physicist says "Let's calculate some cool stuff ..."
$endgroup$
– gandalf61
Sep 11 at 16:33
add a comment
|
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/4.0/"u003ecc by-sa 4.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3352377%2fwhy-arent-tangent-spaces-simply-defined-as-vector-spaces-with-same-dimension-as%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We don't just want to have a vector space to call the "tangent space". We want to do geometric things with the tangent space, and we can't do those things if it's just an arbitrary vector space of the right dimension. We need it to more specifically be a vector space that encodes the parts of the geometry of our manifold that we care about.
For instance, here's one geometric thing we'd like to be able to do: given a smooth curve $gamma:mathbbRto M$ on a manifold $M$, we'd like to define a "tangent vector" at each point of the curve. For each $tinmathbbR$, this should give us a vector "$gamma'(t)$" which is in the tangent space of $M$ at the point $gamma(t)$. If the tangent space is just defined as some arbitrary $n$-dimensional vector space, there isn't going to be any natural way to define $gamma'(t)$. But for the usual definition, there is: we just take the equivalence class of the curve $gamma$.
Another very useful thing we like to do is differentiate functions between manifolds: if $f:Mto N$ is a smooth function between manifolds and $pin M$, there should be a "derivative" $df$ which is a linear map from $T_pMto T_f(p)N$. Again, there's not any natural way to get such a map if the tangent spaces are just some arbitrary vector spaces.
To put it another way, the tangent space to a manifold at a point is not merely a vector space. It is a vector space together with a bunch of extra structure relating it to other geometric features of the manifold. So we care about it not just as a vector space up to isomorphism, but as a "vector space plus extra structure" up to isomorphism. It's possible to axiomatize this extra structure, and then you could define the tangent space as just any abstract gadget satisfying those axioms, rather than restricting yourself to the usual definition by equivalence classes of curves. This isn't commonly done though because (as far as I know) there isn't any nice axiomatization that isn't essentially just saying "a vector space equipped with an isomorphism to this particular concrete construction", so you might as well just use the concrete construction itself.
answered Sep 11 at 2:49
Eric WofseyEric Wofsey
220k15 gold badges256 silver badges399 bronze badges
$endgroup$
$begingroup$
Do you have a reference for axiomatizing this “extra structure” on tangent spaces?
$endgroup$
– Santana Afton
Sep 11 at 4:20
$begingroup$
@SantanaAfton: No, I don't recall ever seeing such a thing written down. Of course, one way to specify the "extra structure" is to just say you have a chosen isomorphism to your favorite specific construction of the tangent space. Any definition is going to be essentially equivalent to this (since of course you want the thing you're defining to be canonically isomorphic to the specific construction), and I don't know of a definition that is really different on its face.
$endgroup$
– Eric Wofsey
Sep 11 at 4:24
$begingroup$
Here's an analogy: you can construct real numbers as Dedekind cuts of rationals. You could also axiomatize real numbers as an ordered field which has exactly one element in each Dedekind cut of the copy of the rationals sitting inside it. But this isn't a particularly useful axiomatization, since it's basically just saying you have an ordered field isomorphic to the ordered field Dedekind cuts of the the rationals. A more useful axiomatization would be that the reals are a complete ordered field.
$endgroup$
– Eric Wofsey
Sep 11 at 4:29
$begingroup$
So, in the case of tangent spaces, it's easy to write down not-useful axiomatizations analogous to that first axiomatization of real numbers, but I don't know of a nice axiomatization analogous the second one. (And if such a thing existed, I suspect it would be used fairly often!)
$endgroup$
– Eric Wofsey
Sep 11 at 4:30
2
$begingroup$
@EricWofsey The definition of the tangent space as equivalence classes of derivations comes to mind, i.e. of operators $D$ on the continuous functions on your manifold that satisfy $D(fg) = fD(g) + D(f)g$. There is no mention of curves or vectors in there so you might be able to spin some substantially different axiomatization from there.
$endgroup$
– mlk
Sep 11 at 12:27
|
show 2 more comments
$begingroup$
We don't just want to have a vector space to call the "tangent space". We want to do geometric things with the tangent space, and we can't do those things if it's just an arbitrary vector space of the right dimension. We need it to more specifically be a vector space that encodes the parts of the geometry of our manifold that we care about.
For instance, here's one geometric thing we'd like to be able to do: given a smooth curve $gamma:mathbbRto M$ on a manifold $M$, we'd like to define a "tangent vector" at each point of the curve. For each $tinmathbbR$, this should give us a vector "$gamma'(t)$" which is in the tangent space of $M$ at the point $gamma(t)$. If the tangent space is just defined as some arbitrary $n$-dimensional vector space, there isn't going to be any natural way to define $gamma'(t)$. But for the usual definition, there is: we just take the equivalence class of the curve $gamma$.
Another very useful thing we like to do is differentiate functions between manifolds: if $f:Mto N$ is a smooth function between manifolds and $pin M$, there should be a "derivative" $df$ which is a linear map from $T_pMto T_f(p)N$. Again, there's not any natural way to get such a map if the tangent spaces are just some arbitrary vector spaces.
To put it another way, the tangent space to a manifold at a point is not merely a vector space. It is a vector space together with a bunch of extra structure relating it to other geometric features of the manifold. So we care about it not just as a vector space up to isomorphism, but as a "vector space plus extra structure" up to isomorphism. It's possible to axiomatize this extra structure, and then you could define the tangent space as just any abstract gadget satisfying those axioms, rather than restricting yourself to the usual definition by equivalence classes of curves. This isn't commonly done though because (as far as I know) there isn't any nice axiomatization that isn't essentially just saying "a vector space equipped with an isomorphism to this particular concrete construction", so you might as well just use the concrete construction itself.
answered Sep 11 at 2:49
Eric WofseyEric Wofsey
220k15 gold badges256 silver badges399 bronze badges
$endgroup$
$begingroup$
Do you have a reference for axiomatizing this “extra structure” on tangent spaces?
$endgroup$
– Santana Afton
Sep 11 at 4:20
$begingroup$
@SantanaAfton: No, I don't recall ever seeing such a thing written down. Of course, one way to specify the "extra structure" is to just say you have a chosen isomorphism to your favorite specific construction of the tangent space. Any definition is going to be essentially equivalent to this (since of course you want the thing you're defining to be canonically isomorphic to the specific construction), and I don't know of a definition that is really different on its face.
$endgroup$
– Eric Wofsey
Sep 11 at 4:24
$begingroup$
Here's an analogy: you can construct real numbers as Dedekind cuts of rationals. You could also axiomatize real numbers as an ordered field which has exactly one element in each Dedekind cut of the copy of the rationals sitting inside it. But this isn't a particularly useful axiomatization, since it's basically just saying you have an ordered field isomorphic to the ordered field Dedekind cuts of the the rationals. A more useful axiomatization would be that the reals are a complete ordered field.
$endgroup$
– Eric Wofsey
Sep 11 at 4:29
$begingroup$
So, in the case of tangent spaces, it's easy to write down not-useful axiomatizations analogous to that first axiomatization of real numbers, but I don't know of a nice axiomatization analogous the second one. (And if such a thing existed, I suspect it would be used fairly often!)
$endgroup$
– Eric Wofsey
Sep 11 at 4:30
2
$begingroup$
@EricWofsey The definition of the tangent space as equivalence classes of derivations comes to mind, i.e. of operators $D$ on the continuous functions on your manifold that satisfy $D(fg) = fD(g) + D(f)g$. There is no mention of curves or vectors in there so you might be able to spin some substantially different axiomatization from there.
$endgroup$
– mlk
Sep 11 at 12:27
|
show 2 more comments
$begingroup$
We don't just want to have a vector space to call the "tangent space". We want to do geometric things with the tangent space, and we can't do those things if it's just an arbitrary vector space of the right dimension. We need it to more specifically be a vector space that encodes the parts of the geometry of our manifold that we care about.
For instance, here's one geometric thing we'd like to be able to do: given a smooth curve $gamma:mathbbRto M$ on a manifold $M$, we'd like to define a "tangent vector" at each point of the curve. For each $tinmathbbR$, this should give us a vector "$gamma'(t)$" which is in the tangent space of $M$ at the point $gamma(t)$. If the tangent space is just defined as some arbitrary $n$-dimensional vector space, there isn't going to be any natural way to define $gamma'(t)$. But for the usual definition, there is: we just take the equivalence class of the curve $gamma$.
Another very useful thing we like to do is differentiate functions between manifolds: if $f:Mto N$ is a smooth function between manifolds and $pin M$, there should be a "derivative" $df$ which is a linear map from $T_pMto T_f(p)N$. Again, there's not any natural way to get such a map if the tangent spaces are just some arbitrary vector spaces.
To put it another way, the tangent space to a manifold at a point is not merely a vector space. It is a vector space together with a bunch of extra structure relating it to other geometric features of the manifold. So we care about it not just as a vector space up to isomorphism, but as a "vector space plus extra structure" up to isomorphism. It's possible to axiomatize this extra structure, and then you could define the tangent space as just any abstract gadget satisfying those axioms, rather than restricting yourself to the usual definition by equivalence classes of curves. This isn't commonly done though because (as far as I know) there isn't any nice axiomatization that isn't essentially just saying "a vector space equipped with an isomorphism to this particular concrete construction", so you might as well just use the concrete construction itself.
answered Sep 11 at 2:49
Eric WofseyEric Wofsey
220k15 gold badges256 silver badges399 bronze badges
$endgroup$
We don't just want to have a vector space to call the "tangent space". We want to do geometric things with the tangent space, and we can't do those things if it's just an arbitrary vector space of the right dimension. We need it to more specifically be a vector space that encodes the parts of the geometry of our manifold that we care about.
For instance, here's one geometric thing we'd like to be able to do: given a smooth curve $gamma:mathbbRto M$ on a manifold $M$, we'd like to define a "tangent vector" at each point of the curve. For each $tinmathbbR$, this should give us a vector "$gamma'(t)$" which is in the tangent space of $M$ at the point $gamma(t)$. If the tangent space is just defined as some arbitrary $n$-dimensional vector space, there isn't going to be any natural way to define $gamma'(t)$. But for the usual definition, there is: we just take the equivalence class of the curve $gamma$.
Another very useful thing we like to do is differentiate functions between manifolds: if $f:Mto N$ is a smooth function between manifolds and $pin M$, there should be a "derivative" $df$ which is a linear map from $T_pMto T_f(p)N$. Again, there's not any natural way to get such a map if the tangent spaces are just some arbitrary vector spaces.
To put it another way, the tangent space to a manifold at a point is not merely a vector space. It is a vector space together with a bunch of extra structure relating it to other geometric features of the manifold. So we care about it not just as a vector space up to isomorphism, but as a "vector space plus extra structure" up to isomorphism. It's possible to axiomatize this extra structure, and then you could define the tangent space as just any abstract gadget satisfying those axioms, rather than restricting yourself to the usual definition by equivalence classes of curves. This isn't commonly done though because (as far as I know) there isn't any nice axiomatization that isn't essentially just saying "a vector space equipped with an isomorphism to this particular concrete construction", so you might as well just use the concrete construction itself.
answered Sep 11 at 2:49
Eric WofseyEric Wofsey
220k15 gold badges256 silver badges399 bronze badges
edited Sep 11 at 4:21
answered Sep 11 at 2:49
Eric WofseyEric Wofsey
220k15 gold badges256 silver badges399 bronze badges
answered Sep 11 at 2:49
Eric WofseyEric Wofsey
220k15 gold badges256 silver badges399 bronze badges
answered Sep 11 at 2:49
Eric WofseyEric Wofsey
220k15 gold badges256 silver badges399 bronze badges
220k15 gold badges256 silver badges399 bronze badges
$begingroup$
Do you have a reference for axiomatizing this “extra structure” on tangent spaces?
$endgroup$
– Santana Afton
Sep 11 at 4:20
$begingroup$
@SantanaAfton: No, I don't recall ever seeing such a thing written down. Of course, one way to specify the "extra structure" is to just say you have a chosen isomorphism to your favorite specific construction of the tangent space. Any definition is going to be essentially equivalent to this (since of course you want the thing you're defining to be canonically isomorphic to the specific construction), and I don't know of a definition that is really different on its face.
$endgroup$
– Eric Wofsey
Sep 11 at 4:24
$begingroup$
Here's an analogy: you can construct real numbers as Dedekind cuts of rationals. You could also axiomatize real numbers as an ordered field which has exactly one element in each Dedekind cut of the copy of the rationals sitting inside it. But this isn't a particularly useful axiomatization, since it's basically just saying you have an ordered field isomorphic to the ordered field Dedekind cuts of the the rationals. A more useful axiomatization would be that the reals are a complete ordered field.
$endgroup$
– Eric Wofsey
Sep 11 at 4:29
$begingroup$
So, in the case of tangent spaces, it's easy to write down not-useful axiomatizations analogous to that first axiomatization of real numbers, but I don't know of a nice axiomatization analogous the second one. (And if such a thing existed, I suspect it would be used fairly often!)
$endgroup$
– Eric Wofsey
Sep 11 at 4:30
2
$begingroup$
@EricWofsey The definition of the tangent space as equivalence classes of derivations comes to mind, i.e. of operators $D$ on the continuous functions on your manifold that satisfy $D(fg) = fD(g) + D(f)g$. There is no mention of curves or vectors in there so you might be able to spin some substantially different axiomatization from there.
$endgroup$
– mlk
Sep 11 at 12:27
|
show 2 more comments
$begingroup$
Do you have a reference for axiomatizing this “extra structure” on tangent spaces?
$endgroup$
– Santana Afton
Sep 11 at 4:20
$begingroup$
@SantanaAfton: No, I don't recall ever seeing such a thing written down. Of course, one way to specify the "extra structure" is to just say you have a chosen isomorphism to your favorite specific construction of the tangent space. Any definition is going to be essentially equivalent to this (since of course you want the thing you're defining to be canonically isomorphic to the specific construction), and I don't know of a definition that is really different on its face.
$endgroup$
– Eric Wofsey
Sep 11 at 4:24
$begingroup$
Here's an analogy: you can construct real numbers as Dedekind cuts of rationals. You could also axiomatize real numbers as an ordered field which has exactly one element in each Dedekind cut of the copy of the rationals sitting inside it. But this isn't a particularly useful axiomatization, since it's basically just saying you have an ordered field isomorphic to the ordered field Dedekind cuts of the the rationals. A more useful axiomatization would be that the reals are a complete ordered field.
$endgroup$
– Eric Wofsey
Sep 11 at 4:29
$begingroup$
So, in the case of tangent spaces, it's easy to write down not-useful axiomatizations analogous to that first axiomatization of real numbers, but I don't know of a nice axiomatization analogous the second one. (And if such a thing existed, I suspect it would be used fairly often!)
$endgroup$
– Eric Wofsey
Sep 11 at 4:30
2
$begingroup$
@EricWofsey The definition of the tangent space as equivalence classes of derivations comes to mind, i.e. of operators $D$ on the continuous functions on your manifold that satisfy $D(fg) = fD(g) + D(f)g$. There is no mention of curves or vectors in there so you might be able to spin some substantially different axiomatization from there.
$endgroup$
– mlk
Sep 11 at 12:27
$begingroup$
Do you have a reference for axiomatizing this “extra structure” on tangent spaces?
$endgroup$
– Santana Afton
Sep 11 at 4:20
$begingroup$
Do you have a reference for axiomatizing this “extra structure” on tangent spaces?
$endgroup$
– Santana Afton
Sep 11 at 4:20
$begingroup$
@SantanaAfton: No, I don't recall ever seeing such a thing written down. Of course, one way to specify the "extra structure" is to just say you have a chosen isomorphism to your favorite specific construction of the tangent space. Any definition is going to be essentially equivalent to this (since of course you want the thing you're defining to be canonically isomorphic to the specific construction), and I don't know of a definition that is really different on its face.
$endgroup$
– Eric Wofsey
Sep 11 at 4:24
$begingroup$
@SantanaAfton: No, I don't recall ever seeing such a thing written down. Of course, one way to specify the "extra structure" is to just say you have a chosen isomorphism to your favorite specific construction of the tangent space. Any definition is going to be essentially equivalent to this (since of course you want the thing you're defining to be canonically isomorphic to the specific construction), and I don't know of a definition that is really different on its face.
$endgroup$
– Eric Wofsey
Sep 11 at 4:24
$begingroup$
Here's an analogy: you can construct real numbers as Dedekind cuts of rationals. You could also axiomatize real numbers as an ordered field which has exactly one element in each Dedekind cut of the copy of the rationals sitting inside it. But this isn't a particularly useful axiomatization, since it's basically just saying you have an ordered field isomorphic to the ordered field Dedekind cuts of the the rationals. A more useful axiomatization would be that the reals are a complete ordered field.
$endgroup$
– Eric Wofsey
Sep 11 at 4:29
$begingroup$
Here's an analogy: you can construct real numbers as Dedekind cuts of rationals. You could also axiomatize real numbers as an ordered field which has exactly one element in each Dedekind cut of the copy of the rationals sitting inside it. But this isn't a particularly useful axiomatization, since it's basically just saying you have an ordered field isomorphic to the ordered field Dedekind cuts of the the rationals. A more useful axiomatization would be that the reals are a complete ordered field.
$endgroup$
– Eric Wofsey
Sep 11 at 4:29
$begingroup$
So, in the case of tangent spaces, it's easy to write down not-useful axiomatizations analogous to that first axiomatization of real numbers, but I don't know of a nice axiomatization analogous the second one. (And if such a thing existed, I suspect it would be used fairly often!)
$endgroup$
– Eric Wofsey
Sep 11 at 4:30
$begingroup$
So, in the case of tangent spaces, it's easy to write down not-useful axiomatizations analogous to that first axiomatization of real numbers, but I don't know of a nice axiomatization analogous the second one. (And if such a thing existed, I suspect it would be used fairly often!)
$endgroup$
– Eric Wofsey
Sep 11 at 4:30
2
2
$begingroup$
@EricWofsey The definition of the tangent space as equivalence classes of derivations comes to mind, i.e. of operators $D$ on the continuous functions on your manifold that satisfy $D(fg) = fD(g) + D(f)g$. There is no mention of curves or vectors in there so you might be able to spin some substantially different axiomatization from there.
$endgroup$
– mlk
Sep 11 at 12:27
$begingroup$
@EricWofsey The definition of the tangent space as equivalence classes of derivations comes to mind, i.e. of operators $D$ on the continuous functions on your manifold that satisfy $D(fg) = fD(g) + D(f)g$. There is no mention of curves or vectors in there so you might be able to spin some substantially different axiomatization from there.
$endgroup$
– mlk
Sep 11 at 12:27
|
show 2 more comments
$begingroup$
If they are “just vector spaces” at each point, then every set would be a manifold.
The whole point is to link the topological structure of the manifold (using coordinate patches that fit together well) to linear algebra that goes on in the tangent spaces.
Another reason to give many different definitions is that they all expose some aspect of what a manifold does. In particular, I’ve found D. Holm’s chapter on manifolds in Geometric mechanics and symmetry very helpful, as it relates several equivalent definitions.
Finally, another contribution to your angst might be the books you’re reading. Some math expositions by physicists can be extremely frustrating for mathematicians, due to different styles in the two disciplines (to put it charitably), so if you’re reading books by such authors it may contribute to what you’re describing.
answered Sep 11 at 2:42
rschwiebrschwieb
116k13 gold badges114 silver badges269 bronze badges
$endgroup$
$begingroup$
+1 for "different styles" point. Very true. Matthew Mathematician says "First assume the following axioms ..."; Peter Physicist says "Let's calculate some cool stuff ..."
$endgroup$
– gandalf61
Sep 11 at 16:33
add a comment
|
$begingroup$
If they are “just vector spaces” at each point, then every set would be a manifold.
The whole point is to link the topological structure of the manifold (using coordinate patches that fit together well) to linear algebra that goes on in the tangent spaces.
Another reason to give many different definitions is that they all expose some aspect of what a manifold does. In particular, I’ve found D. Holm’s chapter on manifolds in Geometric mechanics and symmetry very helpful, as it relates several equivalent definitions.
Finally, another contribution to your angst might be the books you’re reading. Some math expositions by physicists can be extremely frustrating for mathematicians, due to different styles in the two disciplines (to put it charitably), so if you’re reading books by such authors it may contribute to what you’re describing.
answered Sep 11 at 2:42
rschwiebrschwieb
116k13 gold badges114 silver badges269 bronze badges
$endgroup$
$begingroup$
+1 for "different styles" point. Very true. Matthew Mathematician says "First assume the following axioms ..."; Peter Physicist says "Let's calculate some cool stuff ..."
$endgroup$
– gandalf61
Sep 11 at 16:33
add a comment
|
$begingroup$
If they are “just vector spaces” at each point, then every set would be a manifold.
The whole point is to link the topological structure of the manifold (using coordinate patches that fit together well) to linear algebra that goes on in the tangent spaces.
Another reason to give many different definitions is that they all expose some aspect of what a manifold does. In particular, I’ve found D. Holm’s chapter on manifolds in Geometric mechanics and symmetry very helpful, as it relates several equivalent definitions.
Finally, another contribution to your angst might be the books you’re reading. Some math expositions by physicists can be extremely frustrating for mathematicians, due to different styles in the two disciplines (to put it charitably), so if you’re reading books by such authors it may contribute to what you’re describing.
answered Sep 11 at 2:42
rschwiebrschwieb
116k13 gold badges114 silver badges269 bronze badges
$endgroup$
If they are “just vector spaces” at each point, then every set would be a manifold.
The whole point is to link the topological structure of the manifold (using coordinate patches that fit together well) to linear algebra that goes on in the tangent spaces.
Another reason to give many different definitions is that they all expose some aspect of what a manifold does. In particular, I’ve found D. Holm’s chapter on manifolds in Geometric mechanics and symmetry very helpful, as it relates several equivalent definitions.
Finally, another contribution to your angst might be the books you’re reading. Some math expositions by physicists can be extremely frustrating for mathematicians, due to different styles in the two disciplines (to put it charitably), so if you’re reading books by such authors it may contribute to what you’re describing.
answered Sep 11 at 2:42
rschwiebrschwieb
116k13 gold badges114 silver badges269 bronze badges
edited Sep 11 at 2:51
answered Sep 11 at 2:42
rschwiebrschwieb
116k13 gold badges114 silver badges269 bronze badges
answered Sep 11 at 2:42
rschwiebrschwieb
116k13 gold badges114 silver badges269 bronze badges
answered Sep 11 at 2:42
rschwiebrschwieb
116k13 gold badges114 silver badges269 bronze badges
116k13 gold badges114 silver badges269 bronze badges
$begingroup$
+1 for "different styles" point. Very true. Matthew Mathematician says "First assume the following axioms ..."; Peter Physicist says "Let's calculate some cool stuff ..."
$endgroup$
– gandalf61
Sep 11 at 16:33
add a comment
|
$begingroup$
+1 for "different styles" point. Very true. Matthew Mathematician says "First assume the following axioms ..."; Peter Physicist says "Let's calculate some cool stuff ..."
$endgroup$
– gandalf61
Sep 11 at 16:33
$begingroup$
+1 for "different styles" point. Very true. Matthew Mathematician says "First assume the following axioms ..."; Peter Physicist says "Let's calculate some cool stuff ..."
$endgroup$
– gandalf61
Sep 11 at 16:33
$begingroup$
+1 for "different styles" point. Very true. Matthew Mathematician says "First assume the following axioms ..."; Peter Physicist says "Let's calculate some cool stuff ..."
$endgroup$
– gandalf61
Sep 11 at 16:33
add a comment
|
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3352377%2fwhy-arent-tangent-spaces-simply-defined-as-vector-spaces-with-same-dimension-as%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
The tangent space of a smooth, $n$-dimensional manifold is diffeomorphic to $R^n$, but not canonically: to produce such a diffeomorphism, you need to choose coordinates. The utility of the sophisticated definition of the tangent space comes from the fact that it is intrinsic, i.e., does not depend on any auxiliary choices.
$endgroup$
– fmg
Sep 11 at 3:59