Equivalent definition of a tangent space?

Question

I am trying to work through some basic degree theory on manifolds and I found this nice pdf (http://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Bosshardt.pdf) which gets me exactly where I want to go. However, it seems that the tangent space of a manifold $M \subseteq \mathbb{R}^{n}$ has a definition which I have never seen before. Summarizing the text:

1. Let $x \in \mathbb{R}^{n}$. Then a local parametrization near $x$ is a map

$$ \phi: U \rightarrow V $$

such that $U \subseteq \mathbb{R}^{n}$ is open about $0$, $V \subseteq \mathbb{R}^{n}$ is open, and $\phi(0)=x$.

2. Let $M \subseteq \mathbb{R}^{n}$ be a manifold. Then the tangent space at $x \in M$ is defined in the following way: For some parametrization $ \phi: U \rightarrow V $ near $x$, let $\phi_{0}$ be the Jacobian matrix of $\phi$ evaluated at $0$, which can be seen as a linear transformation from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$.Then the tangent space of $M$ at $x$ is defined as $$T_{x}(M)= \phi_{0}(\mathbb{R}^{n})$$

I have always worked with the tangent space as being the set of derivations at a point. I know that there is also an equivalent definition using equivalence classes of curves. However, this seems to simply be a collection of vectors in $\mathbb{R}^{n}$. Again, the tangent space of an $n-$manifold is isomoprhic to $\mathbb{R}^{n}$, but I cannot seem to find any formal equivalence/isomorphism dealing with the formulation given above.

Can anybody point me in the right direction? Thanks!

Answer 1

There are some confused indices in this definition: the $n$ in $M\subseteq\mathbb{R}^n$ does not need to be the same as the dimension of $m$. To avoid confusion, I will use $m$ instead for the dimension of $M$.

To connect this with the tangent vectors as derivations, let us think of $M$ as an abstract manifold and consider the inclusion map $i:M\to\mathbb{R}^n$ as a smooth embedding. For each $p\in M$, $i$ induces an injective linear map on tangent spaces $di_p:T_pM\to T_{i(p)}\mathbb{R}^n$. But the tangent space at any point in $\mathbb{R}^n$ can be canonically identified with $\mathbb{R}^n$, by taking the partial derivatives with respect to each coordinate as a basis for the derivations at each point. So, identifying $T_{i(p)}\mathbb{R}^n$ with $\mathbb{R}^n$, $di_p$ gives an isomorphism between $T_pM$ and some $m$-dimensional linear subspace of $\mathbb{R}^n$. This latter $m$-dimensional linear subspace is what the paper you linked is using as the definition of $T_pM$.

Answer 2

This is only a supplement to Eric Wofsey's answer. I think the paper needs some explanations.

Here are some prelimanaries.

Given a map differentiable map $\psi : U \to V$ between open subsets $U \subset \mathbb R^k$ and $V \subset \mathbb R^l$, we denote by $d\psi(x) : \mathbb R^k \to \mathbb R^l$ the usual derivative of $\psi$ at $x \in U$ which is a linear map represented by the Jacobian matrix. Given a smooth map $\omega : M \to N$ between smooth manifolds, we denote by $T_x\omega : T_x M \to T_{\omega(x)} N$ its derivative in the "manifold sense" which is a linear map between tangent spaces. Many authors also write $d\omega(x)$ instead of $T_x\omega$, but we want to have them distinguished at first glance.

In Definition 2.1 the author defines the concept of a diffeomorphism $f : X \to Y$ between (arbitrary) subsets $X \subset \mathbb R^k$ and $Y \subset \mathbb R^l$. It means that $f$ is a bijection and for each $x \in X$ there exists an open neigborhood $U$ of $x$ in $\mathbb R^k$ and a smooth map $F : U \to \mathbb R^l$ such that $F(\xi) = f(\xi)$ for $\xi \in U \cap X$, similarly for $f^{-1} : Y \to X$. Certainly a diffeomorphism is a homeomorphism, but its local extension will not satisfy $dF(x) \ne 0$ in general.

In Definition 2.2 he defines a subset $X \subset \mathbb R^k$ to be an $n$-dimensional manifold if every $x \in X$ is contained in a set $V \subset X$ open relative to $X$ which is diffeomorphic to an open set $U \subset R^n$. A diffeomorphism $\phi : U \to V$ is called a local parametrization of $X$ near $x$.

What does this say about $\phi$? First, it is easy to see that $\bar \phi = j \circ \phi : U \to \mathbb R^k$ must be smooth, where $j : V \to \mathbb R^k$ denotes inclusion. Second, for each $z \in U$ there exist an open $W \subset \mathbb R^k$ containing $\phi(z)$ and a smooth map $F : W \to \mathbb R^n$ such that $F \mid_{W \cap V} = \phi^{-1} \mid_{W \cap V}$. For a sufficiently small open neigborhood $W'$ of $\phi(z)$ in $W$ we get $F(W') \subset U$. Moreover, for a sufficiently small neighborhood $U'$ of $z$ in $U$ we get $\bar \phi(U') \subset W'$. This shows that $F \circ \bar \phi \mid_{U'} = id$, thus $d\bar \phi(z)$ has rank $n$. This shows that $\bar \phi$ is a smooth embedding. We conclude that $V$ is a smooth submanifold of $\mathbb R^k$ in the usual interpretation. Moreover, $\phi : U \to V$ is diffeomorphism in the manifold sense and with $z = \phi^{-1}(x)$ we have

$$(*) \quad T_z \bar \phi = T_xj \circ T_z\phi: T_z U \to T_{x} \mathbb R^k .$$ Note that $T_z\phi: T_z U \to T_xV$ is an isomorphism and $T_x j : T_xV \to T_{x} \mathbb R^k$ is a linear embedding whose image is an $n$-dimensioanal linear subspace of $T_{x} \mathbb R^k$.

Since there exists a local parametrization of $X$ near any point $x$, we see that $X$ is a smooth submanifold of $\mathbb R^k$.

In particular, there exists a "usual" tangent space $T_x X$ at $x$ which can be canonically identified with $T_x V$.

Now the author assumes that $0 \in U$ and $z= 0$. There are canonical identifications $T_0 U = \mathbb R^n$ and $T_x \mathbb R^k = \mathbb R^k$. Doing so, $T_0 \bar \phi$ is identified with the "Euclidean" derivative $\phi_0 = d \bar \phi(0): \mathbb R^n \to \mathbb R^k$ of $\bar \phi$ at $0$. $\require{AMScd}$ \begin{CD} \mathbb R^n @>{\phi_0}>> \mathbb R^k \\ @A{\approx}AA @A{\approx}AA \\ T_0 U @>{T_0 \bar \phi}>> T_x \mathbb R^k \end{CD}

Now $(*)$ shows that $T_xj$ identifies $T_x V$ with the linear subspace $T_0\bar \phi(T_0 U)$ of $T_x \mathbb R^k$. In other words, we get a canonical identification $$T_x X = \phi_0(\mathbb R^n) \subset \mathbb R^k = T_x \mathbb R^k .$$ \begin{CD} \mathbb R^n @>{\phi_0}>> \phi_0(\mathbb R^n) @>{}>> \mathbb R^k \\ @A{\approx}AA @A{\approx}AA @A{\approx}AA \\ T_0 U @>{T_0 \phi}>> T_xV @>{T_xj}>> T_x \mathbb R^k \end{CD}

$\phi_0(\mathbb R^n)$ can be understood as the "Euclidean tangent space" of the submanifold $X \subset \mathbb R^k$ at $x$. It has a nice geometric interpretation.

The tangent space $T_x M$ can be defined as the set of equivalence classes of $u : (a(u), b(u)) \to M$, where $0 \in (a(u),b(u))$ and $u(0) = x$ ("smooth curves through $x$"). The equivalence relation is given by $u \sim v$ iff they have the same derivative at $0$ with respect to any chart $\varphi : V \to W$, where $V$ is an open neigborhood of $x$ in $M$ and $W \subset \mathbb R^n$ is open. This means that $(\varphi \circ u)'(0) = (\varphi \circ v)'(0)$.

If $U \subset \mathbb R^n$ is open, then we get a canonical isomorphism $T_xU \to \mathbb R^n$ via $[u] \mapsto u'(0)$. Simply take $id_U$ as a chart around $x$.

Now the above $\bar \phi : U \to \mathbb R^l$ embeds $U$ as a the submanifold $V$ of $\mathbb R^l$. Smooth curves $u$ in $U$ through $0$ are mapped to smooth curves $\bar \phi \circ u$ in $\mathbb R^l$ through $x$ whose images are contained in $V$. The set of all $(\bar \phi \circ u)'(0)$ is nothing else than the set of (Euclidean) tangent vectors to the submanifold $V$ at $x$. The set of these vectors is precisely $\bar \phi_0(\mathbb R^k)$.

Answer 3

The following is an argument showing that the definitions of a tangent space as given in "Topology from the Differentiable Viewpoint" by Milnor and the one given in "Introduction to Smooth Manifolds" by Lee are isomorphic. This is essentially my personal summary of the tremendously helpful comments given by Eric Wofsey below.

For a set $X$ and a point $x \in X$, I will define $T_{x}X$ to be the tangent space as given in Lee.

As stated in Milnor, given a smooth manifold $M\subseteq\mathbb{R}^{k}$ and a point $x \in M$, to define the tangent space, we consider some open set $U\subseteq\mathbb{R}^{m}$ and a diffeomorphism $\phi:\,U\rightarrow M$. Assume without loss of generality that $\phi(0)=x$. Then, the tangent space at $x$ is defined to be $D_{0}\phi(\mathbb{R}^{m})$, where $D_{0}\phi$ is the Jacobian of $\phi$ evaluated at $0$, and $D_{0}\phi(\mathbb{R}^{m})$ is the image of the linear transformation $D_{0}\phi:\mathbb{R}^{m}\rightarrow \mathbb{R}^{k}$.

We will show that $D_{0}\phi(\mathbb{R}^{m})$ is isomorphic to $T_{0}M$ in a series of steps below:

1. Because $\phi$ is assumed to be a diffeomorphism, we know that $d\phi_{0}:\,T_{0}U\rightarrow T_{0}M$ is an isomorphism, by Proposition 3.6 in Lee.

2. Since $U$ is open in $\mathbb{R}^{m}$, we know by Proposition 3.9 in Lee that if $i:U\hookrightarrow\mathbb{R}^{m}$ is the inclusion map, then $di_{0}:\,T_{0}U\rightarrow T_{0}\mathbb{R}^{m}$ is an isomorphism.

3. By Proposition 3.2 in Lee, we know that $T_{0}\mathbb{R}^{m}$ is isomorphic to $\mathbb{R}^{m}$.

4. Lastly, because $\phi$ is assumed to be a diffeomorphism, then $\phi^{-1}\circ\phi:U\rightarrow U$ is such that $\phi^{-1}\circ\phi=I_{d}$. Take $0 \in U$ and $y=\phi(0)$. Also, because the subset $M \subseteq \mathbb{R}^{k}$ on which $\phi^{-1}$ is defined need not be open, let $F:\mathbb{R}^{k} \rightarrow \mathbb{R}^{n}$ be a local extension of $\phi^{-1}$ at $y$. Note that it still holds that in a neighborhood around $y$, $F\circ\phi=I_{d}$.

   Then the Jacobian of $F\circ\phi$ at $x$ is given by

$$D_{0}(F\circ\phi)=D_{y}F\circ D_{0}\phi=1_{m\times m},$$ where $D_{0}\phi$ is a $k\times m$ Jacobian matrix and $D_{y}F$ is the $m\times k$ Jacobian matrix. Hence, $D_{0}\phi$ has a left inverse. From standard matrix theory, it follows that $D_{0}\phi$ has rank $m$, and hence $D_{0}\phi(\mathbb{R}^{m})$ is an $m-$ dimensional vector space. Because all $m-$dimensional vector spaces are isomorphic to $\mathbb{R}^{m}$, it follows that $\mathbb{R}^{m}$ and $D_{0}\phi(\mathbb{R}^{m})$ are isomorphic.

Putting 1-4 together, we have $$T_{0}M\overset{(1)}{\cong}T_{0}U\overset{(2)}{\cong}T_{0}\mathbb{R}^{m}\overset{(3)}{\cong}\mathbb{R}^{m}\overset{(4)}{\cong}D_{0}\phi(\mathbb{R}^{m}),$$ completing the argument.