Cotangent Space

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 showing this directly, we show that the dual space of \(T_pM\) is finite-dimensional.

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

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

is well-defined. Then, 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 \(p\in M\), \(T_pM\cong(\mathfrak{m}_p/\mathfrak{m}_p^2)^\ast\) holds.

Proof

First, let an arbitrary \(v\in T_pM\) be given. Restricting it to the subset \(\mathfrak{m}_p\), we have \(v\vert_{\mathfrak{m}_p}\in\Hom_\mathbb{R}(\mathfrak{m}_p,\mathbb{R})\). Now, to show that \(v\vert_{\mathfrak{m}_p}\) gives a well-defined linear map \(\mathfrak{m}_p/\mathfrak{m}_p^2\rightarrow\mathbb{R}\), we must show that \(\mathfrak{m}_p^2\subseteq\ker (v\vert_{\mathfrak{m}_p})\). 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 \(\mathbf{f}_i\mathbf{f}_j\). 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 \(v\) always sends every generator of \(\mathfrak{m}_p^2\) to \(0\), and \(\mathfrak{m}_p^2\subseteq\ker(v\vert_{\mathfrak{m}_p})\); thus to each \(v\in T_pM\) we can associate a suitable element of \((\mathfrak{m}_p/\mathfrak{m}_p^2)^\ast\). 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 \(\mathbf{f}\in \mathcal{C}^\infty_p\). If we let \(\mathbf{f(p)}\) be the constant function with value \(f(p)\), then \(\mathbf{f}-\mathbf{f(p)}\) is an element of \(\mathfrak{m}_p\), and hence defining

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

is well-defined. Since we must show that \(v_L\) defined this way is not only a linear map but also satisfies §Tangent Space, ⁋Definition 3, we can 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 show that \(L\mapsto v_L\) is also a linear map, and furthermore 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 \(T_pM\) is also finite-dimensional, and in this case, since

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

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

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

Proof

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

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

Let \((U,\varphi)\) be a coordinate system centered at \(p\), and write \(\varphi=(x^i)_{i=1}^m\). Since the objects under consideration are all germs, we may shrink \(U\), and thus assume that \(\varphi(U)\) is an open ball centered at the origin. Also, 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

\[g(x)=g(0)+\sum_{i=1}^m\frac{\partial g}{\partial r^i}\bigg\vert_0r^i(x)+\sum_{i,j}r^i(x)r^j(x)\int_0^1(1-t)\frac{\partial^2g}{\partial r^i\partial r^j}\bigg\vert_{tx}\dd{t}\]

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

\[\begin{aligned}f(q)&=f(p)+\sum_{i=1}^m\frac{\partial (f\circ\varphi^{-1})}{\partial r^i}\bigg\vert_0r^i(\varphi(q))+\sum_{i,j}r^i(\varphi(q))r^j(\varphi(q))\int_0^1(1-t)\frac{\partial^2g}{\partial r^i\partial r^j}\bigg\vert_{t\varphi(q)}\dd{t}\\ &=f(p)+\sum_{i=1}^m\frac{\partial(f\circ\varphi^{-1})}{\partial r^i}\bigg\vert_0 x^i(q)+\sum_{i,j} x^i(q)x^j(q)\int_0^1(1-t)\frac{\partial^2 g}{\partial r^i\partial r^j}\bigg\vert_{t\varphi(q)}\dd{t}\end{aligned}\]

Examining the right-hand side, from \(\mathbf{f}\in\mathfrak{m}_p\) we have \(f(p)=0\), and also, as a function of \(q\), the integral on the right-hand side is a \(C^\infty\) function. Now, since the \(x^i\) are all functions satisfying \(x^i(p)=0\), when we pass to germs in the above equation, the double sum on the right-hand side becomes an element of \(\mathfrak{m}_p^2\). Putting all this together,

\[\mathbf{f}=\sum_{i=1}^m\frac{\partial(f\circ\varphi^{-1})}{\partial r^i}\bigg\vert_0\mathbf{x}^i\mod{\mathfrak{m}}_p^2\]

holds. Since \(\mathbf{f}\) is an arbitrary element, 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 \(\varphi(U)\), 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 \(x^i\circ\varphi^{-1}=r^i\), it becomes

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

We have not yet defined \(\partial/\partial x^i\), but we are well acquainted with directional derivatives in Euclidean space. Using this, if we apply \(\partial/\partial r^j\) to both sides of the above equation, the \(0\) on the right-hand side (that is, an element of \(\mathfrak{m}_0^2\)) becomes \(0\) by the Leibniz rule, and hence this equation becomes

\[a_j=\frac{\partial}{\partial r^j}\bigg\vert_0\sum a_i r^i=0\]

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 can obtain not only information about 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 we know well that the tangent space \(T_pM\) is isomorphic to \((\mathfrak{m}_p/\mathfrak{m}_p^2)^\ast\), it seems 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 containing \(p\), \((U,\varphi)\), and for the component functions of \(\varphi\), \(x^i\), the directional derivative in the \(x^i\) direction is the tangent vector defined by the formula

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

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 directions parallel to the coordinate axes defined by the coordinate system.

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

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

To check that this formula is correct, it suffices to apply the tangent vectors on both sides of the above formula to an arbitrary \(\mathbf{f}\) and compare the results. Or, more simply, since tangent vectors send constants and elements of \(\mathfrak{m}_p^2\) to \(0\) and \(\mathbf{f}-f(p)+\mathfrak{m}_p^2\) is expressed as a linear combination of the \(\mathbf{x}^i+\mathfrak{m}_p^2\), it is enough to apply them to \(\mathbf{x}^j\) alone. Applying the right-hand side to \(\mathbf{x}^j\),

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

so the earlier formula holds.

Now that we have defined tangent vectors in rigorous language, from the next post onward, rather than writing elements of \(\mathcal{C}^\infty_p\) as germs \(\mathbf{f}\), we will write them simply as \(f\). 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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