LEC CAL2 9

A multilinear function $T:V^k\to \mathbb{R}$ is called a $k$-tensor on $V$ and the set of all $k$-tensors, denoted ${\mathcal{T}}^k(V)$, is a vector space over $\mathbb{R}$. If $S\in {\mathcal{T}}^k(V)$ and $T\in {\mathcal{T}}^l(V)$, $S\otimes T\in {\mathcal{T}}^{k+k}(V)$ is defined as expected.

Theorem 323.1(Theorem).

Let $v_1,\dots ,v_n$ be a basis for $V$, and let ${\phi }_1,\dots ,{\phi }_n$ be the dual basis, ${\phi }_i(v_j)={\delta }_{ij}$. Then the set of all $k$-fold tensor products

$$\begin{array}{rlr}{\phi }_{i_1}\otimes \cdots \otimes {\phi }_{i_k}& & 1⩽i_1,\dots ,i_k⩽n\end{array}$$

is a basis for ${\mathcal{T}}^k(V)$, which therefore has dimension $n^k$.

011782

If $T\in {\mathcal{T}}^k(V)$ is a $k$-tensor such that

$$T={\ell }_1\otimes \cdots \otimes {\ell }_k,$$

$T$ is called a decomposable $k$-tensor. Note in particular that all the basis elements described in Thm 1 are decomposable. Thus, any $k$-tensor can be written as a linear combination of decomposable $k$-tensors.

Definition 323.2(Pullback).

Let $V$ and $W$ be finite dimensional vector spaces and let $A:V\to W$ be a linear mapping. Define the pullback operation $A^{\ast }:{\mathcal{T}}^k(W)\to {\mathcal{T}}^k(V)$ by $T\mapsto A^{\ast }T$, where $A^{\ast }T:V^k\to \mathbb{R}$ is the function

$$(A^{\ast }T)(v_1,\dots ,v_k):=T(A{v}_1,\dots ,A{v}_k).$$

Note that $A^{\ast }(S\otimes T)=A^{\ast }S\otimes A^{\ast }T$.

Definition 323.3(Definition).

Let $V$ be an $n$-dimensional vector space and $T\in {\mathcal{T}}^k(V)$ a $k$-tensor. For $\sigma \in {\mathfrak{S}}_k$, define $T^{\sigma }\in {\mathcal{T}}^k(V)$ to be

$$T^{\sigma }(v_1,\dots ,v_k):=T(v_{{\sigma }^{-1}(1)},\dots ,v_{{\sigma }^{-1}(k)}).$$

Proposition 323.4(Guillemin & Haine (n.d.) 1.4.12).

1. If $T={\ell }_1\otimes \cdots \otimes {\ell }_k$, ${\ell }_i\in V^{\ast }$, then $T^{\sigma }={\ell }_{\sigma (1)}\otimes \cdots \otimes {\ell }_{\sigma (k)}$.

2. The assignment $T\mapsto T^{\sigma }$ is a linear map ${\mathcal{T}}^k(V)\to {\mathcal{T}}^k(V)$.

3. If $\sigma ,\tau \in {\mathfrak{S}}_k$, then $T^{\sigma \tau }=(T^{\sigma }{)}^{\tau }$.

Proof.

$(1)$

$$\begin{array}{rl}{T}^{\sigma }(v_1,\dots ,v_n)& ={\ell }_1(v_{{\sigma }^{-1}(1)})\cdot \cdots \cdot {\ell }_k(v_{{\sigma }^{-1}(k)})\\ & ={\ell }_{\sigma ({\sigma }^{-1}(1))}(v_{{\sigma }^{-1}(1)})\cdot \cdots \cdot {\ell }_{\sigma ({\sigma }^{-1}(k))}(v_{{\sigma }^{-1}(k)})\\ & ={\ell }_{\sigma (1)}(v_1)\cdot \cdots \cdot {\ell }_{\sigma (k)}(v_k).\end{array}$$

$(3)$

$$\begin{array}{rlr}(T^{\sigma }{)}^{\tau }(v_1,\dots ,v_k)& =(T^{\sigma })(v_{{\tau }^{-1}(1)},\dots ,v_{{\tau }^{-1}(k)})& \\ & =(T^{\sigma })(u_1,\dots ,u_k)& u_i:=v_{{\tau }^{-1}(i)}\\ & =T(u_{{\sigma }^{-1}(1)},\dots ,u_{{\sigma }^{-1}(k)})& \\ & =T(v_{{\tau }^{-1}({\sigma }^{-1}(1))},\dots ,v_{{\tau }^{-1}({\sigma }^{-1}(k))})& \\ & =T(v_{(\sigma \tau {)}^{-1}(1)},\dots ,v_{(\sigma \tau {)}^{-1}(k)})& \\ & =T^{\sigma \tau }(v_1,\dots ,v_k).& \end{array}$$ □

Call a $k$-tensor $T\in {\mathcal{T}}^k(V)$ alternating if $T^{\sigma }=(-1{)}^{\sigma }T$ for all $\sigma \in {\mathfrak{S}}_k$. Denote by ${\Lambda }^k(V)$ the space of all alternating $k$-tensors.

Definition 323.5(Definition).

If $T\in {\mathcal{T}}^k(V)$, define $\text{Alt}(T)$ by

$$\text{Alt}(T):=\frac{1}{k!}\sum _{\sigma \in {\mathfrak{S}}_k}(-1{)}^{\tau }{T}^{\tau }.$$

Proposition 323.6(Properties of$\text{Alt}$).

For $T\in {\mathcal{T}}^k(V)$ and $\sigma \in {\mathfrak{S}}_k$,

1. $(\text{Alt}(T){)}^{\sigma }=(-1{)}^{\sigma }\text{Alt}(T)$.
2. If $T\in {\Lambda }^k(V)$, then $\text{Alt}(T)=T$.
3. $(\text{Alt}(T){)}^{\sigma }=\text{Alt}(T^{\sigma })$
4. The map $\text{Alt}:{\mathcal{T}}^k(V)\to {\mathcal{T}}^k(V)$, $T\mapsto \text{Alt}(T)$ is linear.

If $\omega \in {\Lambda }^k(V)$ and $\eta \in {\Lambda }^l(V)$, then $\omega \otimes \eta$ is not usually in ${\Lambda }^{k+l}(V)$. Define $\omega \wedge \eta \in {\Lambda }^{k+l}(V)$ by

$$\omega \wedge \eta :=\frac{(k+l)!}{k!l!}\text{Alt}(\omega \otimes \eta ).$$

The wedge product has the following properties:

1. $({\omega }_1+{\omega }_2)\wedge \eta ={\omega }_1\wedge \eta +{\omega }_2\wedge \eta$.
2. $\omega \wedge ({\eta }_1+{\eta }_2)=\omega \wedge {\eta }_1+\omega \wedge {\eta }_2$
3. $\alpha \omega \wedge \eta =\omega \wedge \alpha \eta =\alpha (\omega \wedge \eta )$
4. $\omega \wedge \eta =(-1{)}^{kl}\eta \wedge \omega$
5. $A^{\ast }(\omega \wedge \eta )=A^{\ast }(\omega )\wedge A^{\ast }(\eta )$.

$$\begin{array}{rl}{A}^{\ast }(\omega \wedge \eta )(v_1,\dots ,v_{k+l})& =(\omega \wedge \eta )(A{v}_1,\dots ,A{v}_{k+l})\\ & =\frac{(k+l)!}{k!l!}\text{Alt}(\omega \otimes \eta )(A{v}_1,\dots ,A{v}_{k+l})\\ & =\frac{1}{k!l!}\sum _{\sigma \in {\mathfrak{S}}_{k+l}}(-1{)}^{\sigma }(\omega \otimes \eta )(A{v}_{{\sigma }^{-1}(1)},\dots ,A{v}_{{\sigma }^{-1}(k+l)})\\ & =\frac{1}{k!l!}\sum _{\sigma \in {\mathfrak{S}}_{k+l}}(-1{)}^{\sigma }\omega (A{v}_{{\sigma }^{-1}(1)},\dots ,A{v}_{{\sigma }^{-1}(k)})\cdot \eta (A{v}_{{\sigma }^{-1}(k+1)},\dots ,A{v}_{{\sigma }^{-1}(k+l)})\\ & =\frac{1}{k!l!}\sum _{\sigma \in {\mathfrak{S}}_{k+l}}(-1{)}^{\sigma }{A}^{\ast }\omega (v_{{\sigma }^{-1}(1)},\dots ,v_{{\sigma }^{-1}(k)})\cdot A^{\ast }\eta (v_{{\sigma }^{-1}(k+1)},\dots ,v_{{\sigma }^{-1}(k+l)})\\ & =\frac{1}{k!l!}\sum _{\sigma \in {\mathfrak{S}}_{k+l}}(-1{)}^{\sigma }(A^{\ast }\omega \otimes A^{\ast }\eta )(v_{{\sigma }^{-1}(1)},\dots ,v_{{\sigma }^{-1}(k)},v_{{\sigma }^{-1}(k+1)},\dots ,v_{{\sigma }^{-1}(k+l)})\\ & =\frac{(k+l)!}{k!l!}\text{Alt}(A^{\ast }\omega \otimes A^{\ast }\eta )\\ & =A^{\ast }\omega \wedge A^{\ast }\eta .\end{array}$$

Theorem 323.7(Theorem).

1. If $S\in {\mathcal{T}}^k(V)$ and $T\in {\mathcal{T}}^l(V)$ and $\text{Alt}(S)=0$, then $\text{Alt}(S\otimes T)=\text{Alt}(T\otimes S)=0$.
2. $\text{Alt}(\text{Alt}(\omega \otimes \eta )\otimes \theta )=\text{Alt}(\omega \otimes \eta \otimes \theta )=\text{Alt}(\omega \otimes \text{Alt}(\eta \otimes \theta ))$.
3. If $\omega \in {\Lambda }^k(V)$, $\eta \in {\Lambda }^l(V)$, and $\theta \in {\Lambda }^m(V)$, then

$$\begin{array}{r}(\omega \wedge \eta )\wedge \theta =\omega \wedge (\eta \wedge \theta )=\frac{(k+l+m)!}{k!l!m!}\text{Alt}(\omega \otimes \eta \otimes \theta ).\end{array}$$

$(1)$ is proved by breaking the sum over cosets of ${\mathfrak{S}}_k\subseteq {\mathfrak{S}}_{k+l}$.

Theorem 323.8(Theorem).

Let $v_1,\dots ,v_n$ be a basis for $V$ and ${\phi }_1,\dots ,{\phi }_n$ be the dual basis of $V^{\ast }$. The set of all

$$\begin{array}{rlr}{\phi }_{i_1}\wedge \cdots \wedge {\phi }_{i_k}& & 1⩽i_1<i_2<\cdots <i_k⩽n\end{array}$$

is a basis for ${\Lambda }^k(V)$, which therefore has dimension $\left(\begin{array}{c}n\\ k\end{array}\right)$.

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A “map” $\omega$ with $\omega (p)\in {\Lambda }^k({\mathbb{R}}_p^n)$ is called a $k$-form.

If $f:{\mathbb{R}}^n\to \mathbb{R}$ is differentiable, then $Df(p)\in {\Lambda }^1({\mathbb{R}}^n)$. We define the $1$-form $df$ by

$$df(p)(v_p)=Df(p)(v).$$

Every $k$-form $\omega$ can be written as

$$\omega =\sum _{i_1<\cdots <i_k}{\omega }_{i_1,\dots ,i_k}d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}.$$

Theorem 323.9(Theorem).

If $f:{\mathbb{R}}^n\to \mathbb{R}$ is differentiable, then

$$df=D_{1}f\cdot d{x}_1+\cdots +D_{n}f\cdot d{x}_n.$$

Definition 323.10(Pullback).

Let $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$. We have a linear transformation $f_{\ast }:{\mathbb{R}}_p^n\to {\mathbb{R}}_{f(p)}^m$ defined by

$$f_{\ast }(v_p):=(Df(p)(v){)}_{f(p)}.$$

This induces a linear transformation $f^{\ast }:{\Lambda }^k\left({\mathbb{R}}_{f(p)}^m\right)\to {\Lambda }^k\left({\mathbb{R}}_p^n\right)$ as

$$(f^{\ast }A)(v_{1,p},\dots ,v_{k,p})=A(f_{\ast }(v_{1,p}),\dots ,f_{\ast }(v_{k,p})).$$

If $\omega$ is a $k$-form on ${\mathbb{R}}^m$, we can therefore define a $k$-form $f^{\ast }\omega$ on ${\mathbb{R}}^n$ by

$$(f^{\ast }\omega )(p)=f^{\ast }(\omega (f(p))).$$

Theorem 323.11(Theorem).

If $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable, then

\begin{align*} (1) \quad & f^{*}(dx_{i})=\sum_{j=1}^{n} D_{j}f_{i}\cdot dx_{j} \\ (2)\quad & f^{*}(\omega_{1}+\omega_{2} )=f^{*}(\omega_{1})+f^{*}(\omega_{2}) \\ (3)\quad & f^{*}(g\cdot \omega)=(g\circ f)\cdot f^{*}(\omega) \\ (4)\quad & f^{*}(\omega \wedge \eta)=f^{*}\omega \wedge f^{*}\eta. \\ (5) \quad & (g\circ f)^{*}\omega=f^{*}g^{*}\omega.

\end{align*}

> [!Proof]- > > $(1)$ > $$ > \begin{align*} > { f^{*}(dx_{i}) }(p)(v_{p}) & =f^{*}(dx_{i}(f(p)))(v_{p}) \\ > & = dx_{i}(f(p))(f_{*}(v_{p})) \\ > & =dx_{i}(f(p))(Df(p)(v))_{f(p)} \\ > & =\sum_{j=1}^{n} D_{j}f_{i}\cdot v_{j} \\ > & =\sum_{j=1}^{n} D_{j}f_{i}\cdot dx_{j}(p)(v_{p}) .\\ > \end{align*} > $$ > > $(2)$ > $$ > \begin{align*} > f^{*}(\omega_{1}+\omega_{2})(p) & =f^{*}((\omega_{1}+\omega_{2})(f(p))) \\ > & =f^{*}(\omega_{1}(f(p))+\omega_{2}(f(p))) \\ > & =f^{*}(\omega_{1}(f(p)))+f^{*}(\omega_{2}(f(p))) \\ > & =f^{*}(\omega_{1})(p)+f^{*}(\omega_{2})(p) > \end{align*} > $$ > $(3)$ > $$ > \begin{align*} > f^{*}(g\cdot \omega)(p) & =f^{*}((g\cdot \omega)(f(p))) \\ > & =f^{*}(g(f(p))\cdot \omega(f(p))) \\ > & =g(f(p))f^{*}(\omega(f(p))) \\ > & =[(g\circ f)\cdot f^{*}(\omega)](p) > \end{align*} > $$ > $(4)$ > $$ > \begin{align*} > f^{*}(\omega \wedge \eta)(p) & =f^{*}((\omega \wedge \eta)(f(p))) \\ > & =f^{*}(\omega(f(p))\wedge \eta(f(p))) \\ > & =f^{*}(\omega(f(p)))\wedge f^{*}(\eta(f(p))) \\ > & =f^{*}(\omega)(p)\wedge f^{*}(\eta)(p) > \end{align*} > $$

[!Theorem]

1. $d(\omega +\eta )=d\omega +d\eta$.
2. If $\omega$ is a $k$-form and $\eta$ is an $l$-form, then $d(\omega \wedge \eta )=d\omega \wedge \eta +(-1{)}^k\omega \wedge d\eta$.
3. $d(d\omega )=0$. w
4. If $\omega$ is a $k$ form on ${\mathbb{R}}^m$ and $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable, then $f^{\ast }(d\omega )=d(f^{\ast }\omega )$.

[!Proof]-

[!Definition]
A form $\omega$ is called closed if $d\omega =0$ and exact if $\omega =d\eta$ for some $\eta$. exact $⟹$ closed.

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References

Guillemin, V., & Haine, P. J. (n.d.). Differential Forms. Retrieved November 16, 2025, from https://math.mit.edu/classes/18.952/2018SP/files/18.952_book.pdf