10. Differential Forms and Multilinear Algebra

10.1Differential Forms

$M$ is a smooth manifold, $\dim M=n$.

定义 10.1.1. A differential form of degree $k$, or a $k$-form, is a ${\mathcal{C}}^{\infty }(M)$-multilinear map$$\begin{array}{rl}\omega :\mathfrak{X}(M)\times \cdots \times \mathfrak{X}(M)& \to {\mathcal{C}}^{\infty }(M)\\ (X_1,\dots ,X_k)& \mapsto \omega (X_1,\dots ,X_k)\end{array}$$with the property$$\omega (X_{\sigma (1)},\dots ,X_{\sigma (k)})=\mathrm{sign}(\sigma )\cdot \omega (X_1,\dots ,X_k)$$$\mathrm{sign}(\sigma )=\pm 1$ according to whether the number of transpositions in $\sigma$ is even or odd.

引理 10.1.2. $\omega (X_1,\dots ,X_k)(p)$ depends on $X_i$ only through $X_i(p)$ $\Rightarrow$ $\omega (p):T_{p}M\times T_{p}M\to \mathbb{R}$ is $k$-multilinear.$$\omega (p)(X_{\sigma (1)}(p),\dots ,X_{\sigma (k)}(p))=\mathrm{sign}(\sigma )\omega (p)(X_1(p),\dots ,X_k(p)),\omega \in \Gamma ({\Lambda }^k)$$

证明. We only have to prove the Lemma for $i=1$.

Step 1: Suppose there is an open set $U\subset M$, s.t. $X_1{|}_U\equiv 0$. Let $\rho :M\to \mathbb{R}$ be a smooth bump function with $\rho (p)=1$ for a fixed $p\in U$ and $\mathrm{supp}(\rho )\subset U$. Then $\rho \cdot X_1\equiv 0$.$$0=\omega (\rho X_1,X_2,\dots ,X_k)=\rho \cdot \omega (X_1,\dots ,X_k)\Rightarrow \omega (X_1,\dots ,X_k)(p)=0$$

Step 2: Suppose $p\in M$ is such that $X_1(p)=0$. Using a chart $(U,\phi )$ around $p$, we can write$$X_1{|}_U=\sum _{j=1}^n{f}_j\frac{\partial }{\partial x_j},f_j\in {\mathcal{C}}^{\infty }(U)$$Let $\rho :U\to \mathbb{R}$ be a ${\mathcal{C}}^{\infty }$ bump function with $\rho {|}_V\equiv 1$ for $V\subset U$ a smalleer neighborhood of $p$ and $\mathrm{supp}(\rho )\subset U$. Then $\rho \cdot f_j\in {\mathcal{C}}^{\infty }(M)$ and $\rho \cdot f_j{|}_V\equiv f_j$. Similarly, $Y_i=\rho \cdot \frac{\partial }{\partial x_j}\in \mathfrak{X}(M)$ and $Y_j{|}_V\equiv \frac{\partial }{\partial x_j}$. Then $Y:=\sum _{j=1}^n(\rho \cdot f_j)\cdot Y_j\in \mathfrak{X}(M)$ has the property that $Y{|}_V\equiv X_1{|}_V$ $\Rightarrow$ $(X_1-Y){|}_V\equiv 0$, so by Step 1:$$\omega (X_1-Y,X_2,\dots ,X_k)(p)=0$$$$0=\omega (X_1-Y,X_2,\dots ,X_k)(p)=\omega (X_1,\dots ,X_k)(p)-\omega (Y,X_2,\dots ,X_k)(p)$$$$\omega (Y,X_2,\dots ,X_k)(p)=\sum _{j=1}^n\underset{=0,\text{because}{f}_j(p)=0\text{since}{X}_1(p)=0}{\underbrace{(\rho \cdot f_j)(p)}}\omega (Y_j,X_2,\dots ,X_k)(p)=0$$$$\Rightarrow \omega (X_1,\dots ,X_k)(p)=0\text{wherever}{X}_1(p)=0$$

Step 3: Suppose $X_1,X_1^{\prime }\in \mathfrak{X}(M)$ with $X_1(p)=X_1^{\prime }(p)$. Then applying Step 2 to $X_1-X_1^{\prime }$, we see$$\omega (X_1,\dots ,X_k)(p)=\omega (X_1^{\prime },X_2,\dots ,X_k)(p)$$

$\square$

10.2Excursion into Multilinear Algebra

Let $V,W$ be (finite-dimensional) $\mathbb{R}$-vector spaces.

定义 10.2.1. A tensor product for $V$ and $W$ is a bilinear map (Universal property of tensor product), where $T$ is a $\mathbb{R}$-vector space, such that every bilinear map $f:V\times W\to Z$ factorizes uniquely through $\phi$.

定理 10.2.2. A tensor product exists, and is unique up to unique isomorphism.

证明. Uniqueness: Suppose ${\phi }_i:V\times W\to T_i$, $i=1,2$ are two tensor products satisfying the universal property.

$$\begin{array}{rl}{\phi }_2& ={\overline{\phi }}_2\circ {\phi }_1\\ & ={\overline{\phi }}_2\circ {\overline{\phi }}_1\circ {\phi }_2\end{array}\text{and similarly}\begin{array}{rl}{\phi }_1& ={\overline{\phi }}_1\circ {\phi }_2\\ & ={\overline{\phi }}_1\circ {\overline{\phi }}_2\circ {\phi }_1\end{array}$$

Similarly ${\overline{\phi }}_1\circ {\overline{\phi }}_2={\mathrm{Id}}_{T_1}$

$\Rightarrow$ the ${\overline{\phi }}_i$ are isomorphisms inverse to each others.
These are the only choices of isomorphisms between the $T_i$, which make the triangle commute.

Existence: Let $X$ be the $\mathbb{R}$-vector space with basis $V\times W$. Let $Y\subset X$ be the subspace generated by elements in $X$ of the form:$$(a{v}_1+b{v}_2,w)-a(v_1,w)-b(v_2,w)\text{and}(v,a{w}_1+b{w}_2)-a(v,w_1)-b(v,w_2)$$where $T:=X/Y$ is the quotient vector space. The coset of $(v,w)$ will be denoted $v\otimes w$. Define $\begin{array}{rl}\phi :V\times W& \to T\text{by}\\ (v,w)& \mapsto v\otimes w\end{array}$

Claim 10.2.3. $(T,\phi )$ is a tensor porduct of $V$ and $W$.

证明. 1. $\phi$ is bilinear$$\begin{array}{rl}\phi (a{v}_1+b{v}_2,w)& =(a{v}_1+b{v}_2)\otimes w\\ & =a{v}_1\otimes w+a{v}_2\otimes w\end{array}$$So $\phi$ is linear in the first argument. Similar argument for the second argument.

2. Given a bilinear $f:V\times W\to Z$, define $\overline{f}(v\otimes w):=f(v,w)$, and extended linearly to $T$. Then $\overline{f}:T\to Z$ is a well-defined linear map. Moreover, $\overline{f}\circ \phi (v,w)=\overline{f}(v\otimes w)=f(v,w)$, so $f=\overline{f}\circ \phi$.

3. Given $f$, the $\overline{f}$ in $2$ is unique. Suppose $g:T\to Z$ is any linear map with $f=g\circ \phi$. Then$$\overline{f}(v\otimes w)=f(v,w)=g(v\otimes w)$$Since the $v\otimes w$ span $T$, we conclude $\overline{f}\equiv g$.

$\square$

$\square$

From now on, we write $T=V\otimes W$ and $\phi (v,w)=v\otimes w$ for the unique tensor product of $V$ and $W$.

Suppose $v_1,\dots ,v_n\in V$ and $w_1,\dots ,w_m\in W$ are basis of $V$ respectively $W$. Then $v_i\otimes w_j$, $i\in \{1,\dots ,n\}$, $j\in \{1,\dots ,m\}$ is a basis of $V\otimes W$.$$\dim (V\otimes W)=\dim V\cdot \dim W$$

The space of bilinear maps from $V\times W$ to $\mathbb{R}$ is $(V\otimes W{)}^{\ast }$.

If $V_1,\dots ,V_k$ are finite-dimensional $\mathbb{R}$-vector spaces, there is a unique tensor product $V_1\otimes \cdots \otimes V_k$ which has the universal property for $k$-linear maps:

For a single $\mathbb{R}$-vector space $V$ denoted$$T^k(V)=\underset{k\text{factors}}{\underbrace{V\otimes \cdots \otimes V}}$$Let $T^0(V)=\mathbb{R}$ and $T^1(V)=V$, then the tensor algebra of $V$ is$$T(V)=T^{\ast }(V)=\underset{k=0}{\overset{\infty }{⨁}}{T}^k(V)$$The multiplication in this algebra is induced by$$v_1\otimes v_2\otimes \cdots \otimes v_k\cdot w_1\otimes w_2\otimes \cdots \otimes w_l=v_1\otimes \cdots \otimes v_k\otimes w_1\otimes \cdots \otimes w_l,v_i,w_j\in V$$Then $\cdot$ is written $\otimes$ and $T^k(V)\times T^l(V)\stackrel{\otimes }{\to }{T}^{k+1}(V)$.

10.3Exterior Algebra

, where $k$ is multilinear.

Consider only skew-symmetric $f$, so that$$f(v_{\sigma (1)},\dots ,v_{\sigma (k)})=\mathrm{sign}(\sigma )\cdot f(v_1,\dots ,v_k)$$

$$\begin{array}{rl}{T}^{\ast }(V)& =\underset{k\ge 0}{⨁}{T}^k(V)\\ A& \supseteq \text{the ideal generated by }{v}_1\otimes v_2+v_2\otimes v_1\text{ for }{v}_i\in V\text{ (the “alternating ideal”)}\\ \underset{k\ge 0}{⨁}{A}^k& =A\text{where }{A}^k=A\cap T^k(V)\end{array}$$

$$\begin{array}{rl}{A}^0& =0\\ A^1& =0\\ A^2& =\mathrm{span}\{v_1\otimes v_2+v_2\otimes v_1\mid v_i\in V\}\\ & \subseteq T^2(V)=\mathrm{span}\{v_1\otimes v_2\mid v_i\in V\}\\ A^k& =\mathrm{span}\left(\bigcup _{p+q+2=k}{T}^q(V)\otimes A^2\otimes T^p(V)\right)\text{for }k\ge 2\end{array}$$

定义 10.3.1. $T^{\ast }(V)/A:={\Lambda }^{\ast }(V)$ is the exterior algebra of $V$.

Suppose $f$ is skew-symmetric. Then$$f(v_1,v_2)=-f(v_2,v_1)$$$$\begin{array}{rlrlr}\Rightarrow & & \overline{f}(v_1\otimes v_2)=-\overline{f}(v_2\otimes v_1)& & \\ \iff & & \overline{f}(v_1\otimes v_2+v_2\otimes v_1)=0& & \end{array}$$

引理 10.3.2. A $k$-multilinear map $f:V\times \cdots \times V\to Z$ is skew-symmetric if and only if $\overline{f}{|}_{A^k}\equiv 0$.

$f$ is $k$-multilinear and skew-symmetric

$$\Lambda (V)=\underset{k⩾0}{⨁}{\Lambda }^k(V)$$$$\begin{array}{rl}[v_{i_1}\otimes v_{i_2}\otimes \cdots \otimes v_{i_k}]& =v_{i_1}\wedge v_{i_2}\wedge \cdots \wedge v_{i_k}\\ & \text{for }1\le i_1<i_2<\cdots <i_k\le n,\text{ forming a basis for }{\Lambda }^k(V)\end{array}$$Let $\dim V=n$, and $v_1,\dots ,v_n$ is a basis of $V$. Then $v_{i_1}\otimes \cdots \otimes v_{i_k}$, $i_j\in \{1,\dots ,n\}$ form a basis for $T^k(V)$. And $\dim T^k(V)=n^k$.$$[v_{\sigma (1)}\otimes \cdots \otimes v_{\sigma (k)}]=\mathrm{sign}(\sigma )[v_1\otimes \cdots \otimes v_k]\text{in}{\Lambda }^k(V)$$If we have two repeating indices, we are going to have zero, because $v_1\otimes v_1+\cancel{v_1\otimes v_1}\in A^2$. So if you have two indices which are the same, then the corresponding elements in the exterior algebra is zero. For those the indices are different, then you can use this equation to just put them in a sending order, whatever their order have here, up to sign, it is just this. Then we are done.

So we think of the exterior algebra as the quotient of the tensor algebra, we don’t usually write elements in this quotient as cosets this bracket, we just write like this. It says$$\dim {\Lambda }^k(C)=\left(\begin{array}{c}n\\ k\end{array}\right)$$That specify all the spaces. So in particular, $${\Lambda }^k(V)=0\text{if}k>n$$So this graded algebra actually stops after the degree $n$. That was not the case for the tensor. The tensor algebra has arbitrary many elements and tesor algebra has a vector space over $\mathbb{R}$ is infinite dimension. But since the spaces vanish in the degree larger than $n$ for the exterior algebra and these space are finite dimensional, the whole exterior algebra is finite dimension. So$$\dim {\Lambda }^{\ast }(V)=\sum _{k=0}\left(\begin{array}{c}n\\ k\end{array}\right)=(1+1{)}^n=2^n$$

Let us consider something about the induced map of tensor products or exterior products. This is kind of functoriality properties of this instructions. First of all, suppose $f_i:V_i\to W_i$ are linear maps.$$\begin{array}{rl}{V}_1\otimes V_2& \stackrel{f}{\to }{W}_1\otimes W_2\text{is a linear map}\\ v_1\otimes v_2& \mapsto f_1(w_1)\otimes f_2(v_2)\end{array}$$where $v_1\otimes v_2$ are called decomposable elements of $V_1\otimes V_2$. What we do is we’ve constructed the tensor product. It is obviously spaned by these decomposable elements. Then the general element is not decomposable, but it is a linear combination of decomposable elements. Because the decomposable once are spanning set, you can make of this definition. Same thing works for the exterior algebra, if $V_1$ is the same as $V_2$. Using these constructions, every linear map $f:V\to W$ induces an algebra homomorphism$$\begin{array}{rl}T(f):T^{\ast }(V)& \to T^{\ast }(W)\\ v_1\otimes \cdots \otimes v_k& \mapsto f(v_1)\otimes \cdots f(v_k)\end{array}$$Similarly,$$\begin{array}{rl}\Lambda (f):{\Lambda }^{\ast }(V)& \to {\Lambda }^{\ast }(W)\text{is an algebra homomorphism}\\ v_1\wedge \cdots \wedge v_k& \mapsto f(v_1)\wedge \cdots \wedge f(v_k)\end{array}$$

What has this happened to do with determinant? Let $\dim V=n$ and $f:V\to V$ is linear, then$${\Lambda }^n(f):{\Lambda }^n(V)\to {\Lambda }^n(V)$$where ${\Lambda }^n(V)$ is $1$-dimensional.

Claim 10.3.3. ${\Lambda }^n(f)$ is multiplication by $\det (f)$.

where $f$ is $k$-linear nad skew-symmetric. The space of $k$-linear skew-symmetric maps $f:V\times \cdots V\to \mathbb{R}$ is naturally $({\Lambda }^k(V){)}^{\ast }={\Lambda }^k(V^{\ast })$. Then ${\lambda }_1\wedge \cdots \wedge {\lambda }_k\in {\Lambda }^k(V^{\ast })$ acts as a linear map ${\Lambda }^k(V)\to \mathbb{R}$ by $({\lambda }_1\wedge \cdots \wedge {\lambda }_k)(v_1\wedge \cdots \wedge v_k)=\sum _{\sigma }\mathrm{sign}(\sigma ){\lambda }_1(v_{\sigma (1)})\cdots {\lambda }_k(v_{\sigma }(k))\in \mathbb{R}$. For instance,$$\begin{array}{rlrlr}k=2& & ({\lambda }_1\wedge {\lambda }_2)(v_1\wedge v_2)={\lambda }_1(v_1){\lambda }_2(v_2)-{\lambda }_1(v_2){\lambda }_2(v_1)& & \end{array}$$

10.4Multilinear Vector Bundle Theory

If $\omega :\mathfrak{X}(M)\times \cdots \times \mathfrak{X}(M)\to {\mathcal{C}}^{\infty }(M)$ is a differential form of degree $k$ on $M$, then$$\omega (p):T_{p}M\times \cdots \times T_{p}M\to \mathbb{R}$$is well-defined, $k$-linear and skew-symmetric.$$\begin{array}{rlrlr}\Rightarrow & & \omega (p)\in {\Lambda }^k{T}_p^{\ast }M& & \end{array}$$Instead of saying for ${\Lambda }^k{T}^{\ast }M$, more generally, that in fact the multilinear construction we have done are extended from vector spaces to vector spaces. Vector space is a vector bundle over the points. To replace the points by arbitrary manifold, essentially, everything was the same. As an example, we will define tensor for vector bundles. Let $E\stackrel{{\pi }_E}{\to }M$, $F\stackrel{{\pi }_F}{\to }M$ be two smooth vector bundles of rank $k$ and $l$, respectively. We can find an open covering $\{U_i\mid i\in I\}$ of $M$, so that on each $U_i$, both $E$ and $F$ are trivial.$$\begin{array}{rl}{\phi }_i:{\pi }_E^{-1}(U_i)& \to U_i\times {\mathbb{R}}^k\\ {\psi }_i:{\pi }_F^{-1}(U_i)& \to U_i\times {\mathbb{R}}^l\end{array}$$where these are local trivialization. Now the question is do this local definitions fit together properly, you have something is well-defined independently to your local trivialization? $U_i\times ({\mathbb{R}}^k\otimes {\mathbb{R}}^l)$ represents $E\otimes F$ over $U_i$. If $U_i\cap U_j\ne \varnothing$, then$$\begin{array}{rl}{\phi }_j\circ {\phi }_i^{-1}:(U_i\cap U_j)\times {\mathbb{R}}^k& \to (U_i\cap U_j)\times {\mathbb{R}}^k\\ (p,v)& \mapsto (p,g_{ij}(p)\cdot v)\end{array}$$where $g_{ij}:U_i\cap U_j\to G{L}_k(\mathbb{R})$. Similarly,$${\psi }_j\circ {\psi }_i^{-1}(p,v)=(p,f_{ji}(p)\cdot v)$$for smooth $f_{ij}:U_i\cap U_j\to G{L}_l(\mathbb{R})$. Consider $\begin{array}{rl}{g}_{ji}\otimes f_{ji}:U_i\cap U_j& \to G{L}_{k\cdot l}(\mathbb{R})\\ p& \mapsto g_{ji}(p)\otimes f_{ji}(p)\end{array}$ by$$(g_{ji}(p)\otimes f_{ji}(p))(v\otimes w)=g_{ji}(p)(v)\otimes f_{ji}(p)(w)$$where $g_{\ast \ast }\otimes f_{\ast \ast }$ is a cocycle, and $E\otimes F$ is the corresponding vector bundle of rank $k\cdot l$, trivial over each $U_i$. This is how using cocycles to define make precise that the vector bundle $E\otimes F$ is the fibrewise tensor product of the fibres $E$ and $F$. Fibres of $E$ and $F$ form the vector spaces and over every point is just take the tensor product of the fibre.

Now we want to extend this and we are not doing with for the tensor algebra, because tensor algebra is infinite dimension and we don’t want to speak of infinite rank vector bundles.

Given a single vector bundle $E\to M$, we can use this construction to define $T^m(E)\to M$ for every $m⩾0$. This descends to a definition of ${\Lambda }^m(E)\to M$ by taking the quotient bundle $T^m(E)/A^m$.

For example, let $E$, $F$ be vector bundles over $M$, and $f:E\to F$ a homomorphism of vector bundles. Then$$\begin{array}{rl}{T}^m(f):T^m(E)& \to T^m(F)\\ {\Lambda }^m(f):{\Lambda }^m(E)& \to {\Lambda }^m(F)\end{array}$$are also homomorphism of vector bundles. I said the differential forms has a value a the point which is an element of ${\Lambda }^k{T}^{\ast }M$. Now we have constructed this vector bundle and apply this to the cotangent bundle.

$$f^{\ast }{\Lambda }^m(E)={\Lambda }^m(f^{\ast }E)$$

We have now defined ${\Lambda }^k{T}^{\ast }M$, and $\Gamma ({\Lambda }^k{T}^{\ast }M)$ are differential forms of degree $k$ on $M$.

Now we want to apply the above discussion to differential forms. Suppose $f:M\to N$ is a smooth map. Then $f^{\ast }({\Lambda }^k{T}^{\ast }N)$ is a vector bundle over $M$. If $\omega \in \Gamma ({\Lambda }^k{T}^{\ast }N)={\Omega }^k(N)$ is a $k$-form on $N$, then we define $f^{\ast }\omega$ as follows$$(f^{\ast }\omega )(X_1,\dots ,X_k)(p)=\omega (f(p))(D_{p}f(X_1),\dots ,D_{p}f(X_k))$$This is a $k$-form on $M$.

$$\begin{array}{rl}(Df{)}^{\ast }:\underset{=f^{\ast }{T}^{\ast }N}{\underbrace{(f^{\ast }TN{)}^{\ast }}}& \to T^{\ast }N\\ {\Lambda }^k((Df{)}^{\ast }):\underset{=f^{\ast }{\Lambda }^k{T}^{\ast }N}{\underbrace{{\Lambda }^k{f}^{\ast }{T}^{\ast }N}}& \to {\Lambda }^k{T}^{\ast }M\end{array}$$Now we can say the following:$$(f^{\ast }(\omega ))(p)={\Lambda }^k(D_{p}f{)}^{\ast }\omega (f(p))$$where the derivative of $p$ at $f$ is a linear map$$\begin{array}{rl}{D}_{p}f:T_{p}M& \to T_{f(p)}N\\ (D_{p}f{)}^{\ast }:T_{f(p)}^{\ast }N& \to T_p^{\ast }M\end{array}$$Here we doing this not from the cotangent space, but on the exterior products.