Cotangent Space

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Cotangent Space and Dimension

In the previous post we saw that \(T_pM\) is an \(\mathbb{R}\)-vector space. Now we will show that \(T_pM\) is in fact finite-dimensional; instead of proving this directly, we show that the dual space of \(T_pM\) is finite-dimensional.

By §Tangent Space, ⁋Proposition 2, the descending chain of ideals in \(\mathcal{C}^\infty_p\)

\[\mathcal{C}^\infty_p\supset\mathfrak{m}_p\supset\mathfrak{m}_p^2\supset\cdots\]

is well-defined. In particular, we may view \(\mathfrak{m}_p/\mathfrak{m}_p^2\) as a vector space over \(\mathcal{C}^\infty_p/\mathfrak{m}_p\cong\mathbb{R}\).

Lemma 1 For a manifold \(M\) and any point \(p\in M\), we have \(T_pM\cong(\mathfrak{m}_p/\mathfrak{m}_p^2)^\ast\).

Proof

First, let an arbitrary \(v\in T_pM\) be given. Restricting it to the subset \(\mathfrak{m}_p\), we have \(v\vert_{\mathfrak{m}_p}\in\Hom_\mathbb{R}(\mathfrak{m}_p,\mathbb{R})\). To show that \(v\vert_{\mathfrak{m}_p}\) defines a linear map \(\mathfrak{m}_p/\mathfrak{m}_p^2\rightarrow\mathbb{R}\) well, we must show that \(\mathfrak{m}_p^2\subset\ker (v\vert_{\mathfrak{m}_p})\). For a suitable index set \(I\), suppose \(\mathfrak{m}_p\) is the ideal generated by the \(\mathbf{f}_i\). Then \(\mathfrak{m}_p^2\) is the ideal generated by the products \(\mathbf{f}_i\mathbf{f}_j\). But

\[v(\mathbf{f}_i\mathbf{f}_j)=\mathbf{f}_i(p)v(\mathbf{f}_j)+\mathbf{f}_j(p)v(\mathbf{f}_i)=0,\]

so \(v\) sends every generator of \(\mathfrak{m}_p^2\) to \(0\); hence \(\mathfrak{m}_p^2\subset\ker(v\vert_{\mathfrak{m}_p})\), and therefore for each \(v\in T_pM\) we can associate a suitable element of \(\mathfrak{m}_p/\mathfrak{m}_p^2\). That this correspondence is an \(\mathbb{R}\)-linear map is obvious.

Conversely, let an arbitrary \(L\in(\mathfrak{m}_p/\mathfrak{m}_p^2)^\ast\) be given, and let us use it to construct a tangent vector \(v_L\). A tangent vector is a linear map from \(C_p^\infty\) to \(\mathbb{R}\), so constructing \(v_L\) amounts to specifying the value of \(v_L(\mathbf{f})\) for each \(\mathbf{f}\in \mathcal{C}^\infty_p\). Let \(\mathbf{f(p)}\) denote the constant function with value \(f(p)\); then \(\mathbf{f}-\mathbf{f(p)}\) is an element of \(\mathfrak{m}_p\), and hence

\[v_L(\mathbf{f})=L((\mathbf{f}-\mathbf{f(p)})+\mathfrak{m}_p^2)\]

is well-defined. We must show that this \(v_L\) is not only a linear map but also satisfies §Tangent Space, ⁋Definition 3; thus we compute

\[\begin{aligned} v_L(\mathbf{f}\cdot\mathbf{g})&=L((\mathbf{fg}-\mathbf{f(p)g(p)})+\mathfrak{m}_p^2)\\ &=L(((\mathbf{f}-\mathbf{f(p)})(\mathbf{g}-\mathbf{g(p)})+\mathbf{f(p)}(\mathbf{g}-\mathbf{g(p)})+(\mathbf{f}-\mathbf{f(p)})\mathbf{g(p)})+\mathfrak{m}_p^2)\\ &=\mathbf{f}(p)L((\mathbf{g}-\mathbf{g(p)})+\mathfrak{m}_p^2)+\mathbf{g}(p)L((\mathbf{f}-\mathbf{f(p)})+\mathfrak{m}_p^2)\\ &=\mathbf{f}(p)+v_L(\mathbf{g})+\mathbf{g}(p)v_L(\mathbf{f}). \end{aligned}\]

It is easy to see that \(L\mapsto v_L\) is also a linear map, and moreover one can verify that this correspondence is the inverse of the linear map from \(T_pM\) to \((\mathfrak{m}_p/\mathfrak{m}_p^2)^\ast\) defined above.

Therefore, if \(\mathfrak{m}_p/\mathfrak{m}_p^2\) is finite-dimensional then so is \(T_pM\), and in this case

\[(T_pM)^\ast\cong(\mathfrak{m}_p/\mathfrak{m}_p^2)^{\ast\ast}\cong\mathfrak{m}_p/\mathfrak{m}_p^2\]

so we call \(\mathfrak{m}_p/\mathfrak{m}_p^2\) the cotangent space.

Theorem 2 The \(\mathbb{R}\)-vector space \(\mathfrak{m}_p/\mathfrak{m}_p^2\) is finite-dimensional, and its dimension equals that of the manifold \(M\).

Proof

To show this, we use the following multivariate Taylor approximation:

\[\begin{aligned}g(x)&=g(x_0)+\sum_{i=1}^m\frac{\partial g}{\partial r^i}\bigg|_{x_0}(r^i(x)-r^i(x_0))\\ &\phantom{phantom}+\sum_{i,j}(r^i(x)-r^i(x_0))(r^j(x)-r^j(x_0))\int_0^1(1-t)\frac{\partial^2g}{\partial r^i\partial r^j}\bigg|_{(x_0+t(x-x_0))}\mathop{dt}\end{aligned}\]

Let \((U,\varphi)\) be a coordinate system centered at \(p\), and write \(\varphi=(x^i)_{i=1}^m\). Let \(\mathbf{f}\in\mathfrak{m}_p\) be given arbitrarily.

Since the above formula holds in Euclidean space, set \(g=f\circ\varphi^{-1}\) and regard the domain of \(g\) as \(\varphi(U)\). From the Taylor approximation centered at the origin, for any \(x\in\varphi(U)\) we obtain

\[g(x)=g(0)+\sum_{i=1}^m\frac{\partial g}{\partial r^i}\bigg|_0r^i(x)+\sum_{i,j}r^i(x)r^j(x)\int_0^1(1-t)\frac{\partial^2g}{\partial r^i\partial r^j}\bigg|_{tx}\mathop{dt}\]

Now set \(x=\varphi(q)\); then

\[\begin{aligned}f(q)&=f(p)+\sum_{i=1}^m\frac{\partial (f\circ\varphi^{-1})}{\partial r^i}\bigg|_0r^i(\varphi(q))+\sum_{i,j}r^i(\varphi(q))r^j(\varphi(q))\int_0^1(1-t)\frac{\partial^2g}{\partial r^i\partial r^j}\bigg|_{t\varphi(q)}\mathop{dt}\\ &=f(p)+\sum_{i=1}^m\frac{\partial(f\circ\varphi^{-1})}{\partial r^i}\bigg|_0 x^i(q)+\sum_{i,j} x^i(q)x^j(q)\int_0^1(1-t)\frac{\partial^2 g}{\partial r^i\partial r^j}\bigg|_{t\varphi(q)}dt\end{aligned}\]

Examining the right-hand side, since \(\mathbf{f}\in\mathfrak{m}_p\) we have \(f(p)=0\), and the integral term is a \(C^\infty\) function of \(q\). The \(x^i\) are functions satisfying \(x^i(p)=0\), so passing to germs, the double sum on the right-hand side becomes an element of \(\mathfrak{m}_p^2\). Collecting everything,

\[\mathbf{f}=\sum_{i=1}^m\frac{\partial(f\circ\varphi^{-1})}{\partial r^i}\bigg|_0\mathbf{x}^i\mod \mathfrak{m}_p^2\]

holds. Since \(\mathbf{f}\) is arbitrary, we see that \(\mathfrak{m}_p/\mathfrak{m}_p^2\) is generated by the \(\mathbf{x}^i+\mathfrak{m}_p^2\).

To complete the proof we must show that these \(m\) elements \(\mathbf{x}^i+\mathfrak{m}_p^2\) are linearly independent. Suppose

\[\sum_{i=1}^m a_i\mathbf{x}^i=0\mod \mathfrak{m}_p^2\]

Then on \(U\) this equation becomes

\[\sum_{i=1}^m a_i (x^i\circ\varphi^{-1})=0\mod \mathfrak{m}_0^2\]

(where \(\mathfrak{m}_0\) is the maximal ideal corresponding to the point \(0\in\varphi(U)\)), and since \(x^i\circ\varphi^{-1}=r^i\),

\[\sum_{i=1}^m a_i\mathbf{r}^i=0\mod \mathfrak{m}^2_0\]

We have not yet defined \(\partial/\partial x^i\), but we are well acquainted with directional derivatives in Euclidean space. Applying \(\partial/\partial r^j\) to both sides of the above equation, the \(0\) on the right-hand side (that is, some element of \(\mathfrak{n}^2\)) becomes \(0\) by the Leibniz rule, and hence this equation yields

\[a_j=\frac{\partial}{\partial r^j}\bigg|_0\sum a_i r^i=0\]

Therefore \(a_j=0\) holds for all \(j\), and the \(\mathbf{x}^i+\mathfrak{m}_p^2\) are linearly independent.

Looking carefully at the proof of this theorem, we obtain not only the dimension but also a basis of \(T_pM\). We showed that the \(\mathbf{x}^i+\mathfrak{m}_p^2\) form a basis of \(\mathfrak{m}_p/\mathfrak{m}_p^2\), and since the tangent space \(T_pM\) is isomorphic to \((\mathfrak{m}_p/\mathfrak{m}_p^2)^\ast\), it is natural to take their dual basis as a basis of \(T_pM\). That is,

Definition 3 Let a manifold \(M\) and \(p\in M\) be given. For a coordinate system \((U,\varphi)\) containing \(p\) and the component functions \(x^i\) of \(\varphi\), the directional derivative in the \(x^i\) direction is the tangent vector defined by the formula

\[\left(\frac{\partial}{\partial x^i}\bigg|_p\right)f=\frac{\partial(f\circ \varphi^{-1})}{\partial r^i}\bigg|_{\varphi(p)}\]

Of course all tangent vectors can be thought of as directional derivatives, but what is special about these vectors \(\partial/\partial x^i\) is that they are exactly the derivatives in the directions parallel to the coordinate axes determined by the coordinate system.

These \(m\) tangent vectors are, as noted above, the dual basis of \(\mathfrak{m}_p/\mathfrak{m}_p^2\), so it is obvious that they form a basis. Hence, for any \(v\in T_pM\), the expression of \(v\) as a linear combination of the \(\partial/\partial x^i\) must also exist uniquely. It is given by the formula

\[v=\sum_{i=1}^m v(x^i)\frac{\partial}{\partial x^i}\bigg|_p\]

To check that this formula is correct, it suffices to apply the tangent vectors on both sides to an arbitrary \(\mathbf{f}\) and compare the results. Or, more simply, since \(\mathbf{f}\) is represented as a linear combination of the \(\mathbf{x}^i+\mathfrak{m}_p^2\), it is enough to test on \(\mathbf{x}^j\) alone. Applying the right-hand side to \(\mathbf{x}^j\),

\[\sum_{i=1}^m v(x^i)\frac{\partial}{\partial x^i}\bigg|_p x^j=\sum_{i=1}^m v(x^i)\delta_{ij}=v(x^j)\]

so the formula holds.

Now that we have defined tangent vectors in rigorous language, from the next post onward we will write elements of \(\mathcal{C}^\infty_p\) simply as \(f\) rather than as germs \(\mathbf{f}\). Here \(f\in \mathcal{C}^\infty_p\) means that \(f\) is a \(C^\infty\) function defined on a suitable open neighborhood of \(p\in M\).

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References

[War] Frank W. Warner. Foundations of Differentiable Manifolds and Lie Groups, Graduate texts in mathematics, Springer, 2013
[Lee] John M. Lee. Introduction to Smooth Manifolds, Graduate texts in mathematics, Springer, 2012

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