# Why aren't tangent spaces simply defined as vector spaces with same dimension as the manifold?

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 $$\mathbb{R}^n$$, why do we bother with all the long definitions involving derivations and equivalence classes of curves in the first place?

• 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. – fmg Sep 11 at 3:59

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 $$\gamma:\mathbb{R}\to M$$ on a manifold $$M$$, we'd like to define a "tangent vector" at each point of the curve. For each $$t\in\mathbb{R}$$, 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 $$f:M\to N$$ is a smooth function between manifolds and $$p\in M$$, there should be a "derivative" $$df$$ which is a linear map from $$T_pM\to T_{f(p)}N$$. 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.

• Do you have a reference for axiomatizing this “extra structure” on tangent spaces? – Santana Afton Sep 11 at 4:20
• @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. – Eric Wofsey Sep 11 at 4:24
• 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. – Eric Wofsey Sep 11 at 4:29
• 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!) – Eric Wofsey Sep 11 at 4:30
• @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. – mlk Sep 11 at 12:27

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.

• +1 for "different styles" point. Very true. Matthew Mathematician says "First assume the following axioms ..."; Peter Physicist says "Let's calculate some cool stuff ..." – gandalf61 Sep 11 at 16:33