# Math 31AH: Lecture 22

Let us now use the symmetric and antisymmetric tensor products to define two subspaces of the tensor square $\mathbf{V} \otimes \mathbf{V}$ which store “unevaluated” symmetric and antisymmetric tensor products of vectors from $\mathbf{V}.$ The symmetric square of $\mathbf{V}$ is the subspace $\mathbf{V} \vee \mathbf{V}$ of $\mathbf{V} \otimes \mathbf{V}$ spanned by all symmetric tensor products

$\mathbf{v}_1 \vee \mathbf{v}_2, \quad \mathbf{v}_1,\mathbf{v}_2 \in \mathbf{V}.$

Elements of $\mathbf{V} \vee \mathbf{V}$ are called symmetric tensors. Similarly, the antisymmetric square of $\mathbf{V}$ is the subspace $\mathbf{V} \wedge \mathbf{V}$ of $\mathbf{V} \otimes \mathbf{V}$ spanned by all antisymmetric tensor products,

$\mathbf{v}_1 \wedge \mathbf{v}_2, \quad \mathbf{v}_1,\mathbf{v}_2 \in \mathbf{V}.$

Elements of $\mathbf{V} \vee \mathbf{V}$ are called antisymmetric tensors.

All of what we have said above can be generalized in a natural way to products of more than two vectors. More precisely, for any natural number $\mathbf{d} \in \mathbb{N},$ we can define the $d$th tensor power of the vector space $\mathbf{V}$ to be the new vector space $\mathbf{V}^{\otimes d}$ spanned by all “unevaluated” products

$\mathbf{v}_1 \otimes \dots \otimes \mathbf{v}_d$

of $d$ vectors $\mathbf{v}_1,\dots,\mathbf{v}_d.$ The only feature of such multiple unevaluated products is that they are “multilinear,” which really just means that they behave like ordinary products (sans commutativity). For example, in the case $d=3,$ this just means that we have the following three identities in the vector space $\mathbf{V}^{\otimes 3}$: for any scalars $a_1,a_2 \in \mathbb{R}$

$(a_1\mathbf{u}_1 + a_2\mathbf{u}_2) \otimes \mathbf{v} \otimes \mathbf{w} = a_1\mathbf{u}_1 \otimes \mathbf{v} \otimes \mathbf{w} + a_2\mathbf{u}_2 \otimes \mathbf{v} \otimes \mathbf{w}$

for all $\mathbf{u}_1,\mathbf{u}_2,\mathbf{v},\mathbf{w} \in \mathbf{V},$ and

$\mathbf{u} \otimes (a_1\mathbf{v}_1 + a_2\mathbf{v}_2) \otimes \mathbf{w} = a_1 \mathbf{u} \otimes \mathbf{v}_1 \otimes \mathbf{w} + a_2\mathbf{u} \otimes \mathbf{v}_2 \mathbf{w}$

for all $\mathbf{u},\mathbf{v}_1,\mathbf{v}_2,\mathbf{w} \in \mathbf{V},$ and

$\mathbf{u} \otimes \mathbf{v} \otimes (a_1\mathbf{w}_1 + a_2\mathbf{w}_2) = a_1 \mathbf{u} \otimes \mathbf{v} \otimes \mathbf{w}_1 + a_2 \mathbf{u} \otimes \mathbf{v} \otimes \mathbf{w}_2$

for all $\mathbf{u},\mathbf{v},\mathbf{w}_1,\mathbf{w}_2 \in \mathbf{V}.$ If $\mathbf{V}$ comes with a scalar product $\langle \cdot,\cdot \rangle,$ we can use this to define a scalar product on $\mathbf{V}^{\otimes d}$ in a very simple way by declaring

$\langle \mathbf{v}_1 \otimes \dots \otimes \mathbf{v}_d,\mathbf{w}_1 \otimes \dots \otimes \mathbf{w}_d \rangle = \langle \mathbf{v}_1,\mathbf{w}_1 \rangle \dots \langle \mathbf{v}_d. \mathbf{w}_d\rangle.$

Even better, we can use the scalar product so defined to construct an orthonormal basis of $\mathbf{V}^{\otimes d}$ from a given orthonormal basis $\mathbf{E}=\{\mathbf{e}_1,\dots,\mathbf{e}_n\}:$ such a basis is simply given by all tensor products with $d$ factors such that each factor is a vector in $\mathbf{V}.$ More precisely, these are the tensors

$\mathbf{e}_{i(1)} \otimes \mathbf{e}_{i(2)} \otimes \dots \otimes \mathbf{e}_{i(d)}, \quad i \in \mathrm{Fun}(d,N),$

where $\mathrm{Fun}(d,N)$ is a fun notation for the set of all functions

$i \colon \{1,\dots,d\} \to \{1,\dots,N\}.$

In particular, since the cardinality of $\mathrm{Fun}(d,N)$ is $N^d$ (make $N$ choices $d$ times), the dimension of the vector space $\mathbf{V}^{\otimes d}$ is $N^d.$

Example 1: If $\mathbf{V}$ is a $2$-dimensional vector space with orthonormal basis $\{\mathbf{e}_1,\mathbf{e}_2\},$ then an orthonormal basis of $\mathbf{V}^{\otimes 3}$ is given by the tensors

$\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1, \\ \mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_2, \mathbf{e}_1 \otimes \mathbf{e}_2 \otimes \mathbf{e}_1,\mathbf{e}_2 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1, \\ \mathbf{e}_1 \otimes \mathbf{e}_2 \otimes \mathbf{e}_2, \mathbf{e}_2 \otimes \mathbf{e}_1 \otimes \mathbf{e}_2, \mathbf{e}_2 \otimes \mathbf{e}_2 \otimes \mathbf{e}_1, \\ \mathbf{e}_2 \otimes \mathbf{e}_2 \otimes \mathbf{e}_2.$

We now define the $d$-fold symmetric and antisymmetric tensor products. These products rely on the concept of permutations.

Reading Assignment: Familiarize yourself with permutations. What is important for our purposes is that you understand how to multiply permutations, and that you understand what the sign of a permutation is. Feel free to ask questions as needed.

Definition 1: For any $d \in \mathbb{N},$ and any vectors $\mathbf{v}_1,\dots,\mathbf{v}_d \in \mathbf{V},$ we define the symmetric tensor product of these vectors by

$\mathbf{v}_1 \vee \dots \vee \mathbf{v}_d = \frac{1}{d!} \sum\limits_{\pi \in \mathrm{S}(d)} \mathbf{v}_{\pi(1)} \otimes \dots \otimes \mathbf{v}_{\pi(d)},$

and denote by $\mathbf{V}^{\vee d}$ the subspace of $\mathbf{V}^{\otimes d}$ spanned by all symmetric tensor products of $d$ vectors from $\mathbf{V}.$ Likewise, we define the antisymmetric tensor product of $\mathbf{v}_1,\dots,\mathbf{v}_d$ by

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d = \frac{1}{d!} \sum\limits_{\pi \in \mathrm{S}(d)} \mathrm{sgn}(\pi)\mathbf{v}_{\pi(1)} \otimes \dots \otimes \mathbf{v}_{\pi(d)},$

and denote by $\mathbf{V}^{\wedge d}$ the subspace of $\mathbf{V}^{\otimes d}$ spanned by all antisymmetric tensor products of $d$ vectors from $\mathbf{V}.$

Note that, in the case $d=2,$ this definition coincides with the definitions

$\mathbf{v}_1 \vee \mathbf{v}_2 = \frac{1}{2}\left( \mathbf{v}_1\otimes \mathbf{v}_2 + \mathbf{v}_2 \otimes \mathbf{v}_1\right)$

and

$\mathbf{v}_1 \wedge \mathbf{v}_2 = \frac{1}{2}\left(\mathbf{v}_1\otimes \mathbf{v}_2 - \mathbf{v}_2 \otimes \mathbf{v}_1\right)$

from Lecture 21.

Since the symmetric and antisymmetric tensor products are defined in terms of the tensor product, they inherit multilinearity. For example, in the case $d=3,$ this means that we have the following three identities in the vector space $\mathbf{V}^{\vee 3}$: for any scalars $a_1,a_2 \in \mathbb{R}$

$(a_1\mathbf{u}_1 + a_2\mathbf{u}_2) \vee \mathbf{v} \vee \mathbf{w} = a_1\mathbf{u}_1 \vee \mathbf{v} \vee \mathbf{w} + a_2\mathbf{u}_2 \vee \mathbf{v} \vee \mathbf{w}$

for all $\mathbf{u}_1,\mathbf{u}_2,\mathbf{v},\mathbf{w} \in \mathbf{V},$ and

$\mathbf{u} \vee (a_1\mathbf{v}_1 + a_2\mathbf{v}_2) \vee \mathbf{w} = a_1 \mathbf{u} \vee \mathbf{v}_1 \vee \mathbf{w} + a_2\mathbf{u} \vee \mathbf{v}_2 \mathbf{w}$

for all $\mathbf{u},\mathbf{v}_1,\mathbf{v}_2,\mathbf{w} \in \mathbf{V},$ and

$\mathbf{u} \vee \mathbf{v} \vee (a_1\mathbf{w}_1 + a_2\mathbf{w}_2) = a_1 \mathbf{u} \vee \mathbf{v} \vee \mathbf{w}_1 + a_2 \mathbf{u} \vee \mathbf{v} \vee \mathbf{w}_2$

for all $\mathbf{u},\mathbf{v},\mathbf{w}_1,\mathbf{w}_2 \in \mathbf{V}.$ The analogous statements hold in $\mathbf{V}^{\wedge 3}.$

The symmetric tensor product is constructed in such a way that

$\mathbf{v}_{\pi(1)} \vee \dots \vee \mathbf{v}_{\pi(d)} = \mathbf{v}_1 \vee \dots \vee \mathbf{v}_d$

for any permutation $\pi \in \mathrm{S}(d),$ whereas the antisymmetric tensor product is constructed in such a way that

$\mathbf{v}_{\pi(1)} \wedge \dots \wedge \mathbf{v}_{\pi(d)} = \mathrm{sgn}(\pi)\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d$

for any permutation $\pi \in \mathrm{S}(d).$ In particular, if any two of the vectors $\mathbf{v}_1,\dots,\mathbf{v}_d$ are equal, then

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d = \mathbf{0}.$

Indeed, suppose that $\mathbf{v}_1=\mathbf{v}_2.$ On one hand, by the above antisymmetry we have

$\mathbf{v}_2 \wedge \mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d = - \mathbf{v}_1 \wedge \mathbf{v}_2 \wedge \dots \wedge \mathbf{v}_d,$

but on the other hand we also have

$\mathbf{v}_2 \wedge \mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d = \mathbf{v}_1 \wedge \mathbf{v}_2 \wedge \dots \wedge \mathbf{v}_d$

because $\mathbf{v}_1=\mathbf{v}_2.$ This means that

$\mathbf{v}_1 \wedge \mathbf{v}_2 \wedge \dots \wedge \mathbf{v}_d = - \mathbf{v}_1 \wedge \mathbf{v}_2 \wedge \dots \wedge \mathbf{v}_d$

if $\mathbf{v}_1=\mathbf{v}_2,$ which forces

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d = \mathbf{0}.$

The vector space $\mathbf{V}^{\vee d}$ is called the $d$th symmetric power of $\mathbf{V},$ and its elements are called symmetric tensors of degree $d.$ The vector space $\mathbf{V}^{\wedge d}$ is called the $d$th antisymmetric power of $\mathbf{V},$ and its elements are called antisymmetric tensors of degree $d.$ These vector spaces have a physical interpretation. In quantum mechanics, an $n$-dimensional vector space $\mathrm{dim} \mathbf{V}$ is viewed as the state space of a particle that can be in any one of $n$ quantum states. The space $\mathbf{V}^{\vee d}$ is then the state space of $d$ bosons, each of which may occupy one of $n$ quantum states, while $\mathbf{V}^{\wedge d}$ is the state space of $d$ fermions, each of which may be in any of $n$ quantum states. The vanishing of wedge products with two equal factors corresponds physically to the characteristic feature of fermions, i.e. the Pauli exclusion principle. You don’t have to know any of this — I included this perspective in order to provide some indication that the construction of these vector spaces is not just abstract nonsense.

Theorem 1: For any $\mathrm{d} \in \mathbb{N}$ and any $\mathbf{v}_1,\dots,\mathbf{v}_d \in \mathbf{V},$ we have

$\langle \mathbf{v}_1 \vee \dots \vee \mathbf{v}_d,\mathbf{w}_1 \vee \dots \vee \mathbf{w}_d \rangle = \frac{1}{d!} \sum\limits_{\pi \in \mathrm{S}(d)} \langle \mathbf{v}_1,\mathbf{w}_{\pi(1)}\rangle \dots \langle \mathbf{v}_d,\mathbf{w}_{\pi(d)}\rangle,$

and

$\langle \mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d,\mathbf{w}_1 \wedge \dots \wedge \mathbf{w}_d \rangle = \frac{1}{d!} \sum\limits_{\pi \in \mathrm{S}(d)} \mathrm{sgn}(\pi)\langle \mathbf{v}_1,\mathbf{w}_{\pi(1)}\rangle \dots \langle \mathbf{v}_d,\mathbf{w}_{\pi(d)}\rangle.$

Since we won’t use this theorem much, we will skip the proof. However, the proof is not too difficult, and is an exercise in permutations: simply plug in the definitions of the symmetric and antisymmetric tensor products in terms of the original tensor products, expand the scalar product, and simplify.

Perhaps counterintuitively, the antisymmetric tensor product is more important than the symmetric tensor product in linear algebra. The next theorem explains why.

Theorem 2: For any $d \in \mathbb{N}$ and any $\mathbf{v}_1,\dots,\mathbf{v}_d,$ the set $\{\mathbf{v}_1,\dots,\mathbf{v}_d\}$ is linearly dependent if and only if

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d = \mathbf{0}.$

Proof: Suppose first that $\{\mathbf{v}_1,\dots,\mathbf{v}_d\}$ is a linearly dependent set of vectors in $\mathbf{V}.$ If $d=1,$ this means that $\mathbf{v}_1=\mathbf{0},$ whence

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d = \mathbf{v}_1 = \mathbf{0}.$

If $d \geq 2,$ then without loss in generality, the vector $\mathbf{v}_1$ is a linear combination of the vectors $\mathbf{v}_2,\dots,\mathbf{v}_d,$

$\mathbf{v}_1 = a_2\mathbf{v}_2 + \dots + a_d\mathbf{v}_d.$

We then have that

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d = \left(\sum\limits_{i=2}^d a_i\mathbf{v}_i \right) \wedge \mathbf{v}_2 \wedge \dots \wedge \mathbf{v}_d = \sum\limits_{i=2}^d a_i\mathbf{v}_i \wedge \mathbf{v}_2 \wedge \dots \wedge \mathbf{v}_d,$

by multinearity of the wedge product. Now observe that the $i$th term in the sum is a scalar multiple of the wedge product

$\mathbf{v}_i \wedge \dots \wedge \mathbf{v}_i \wedge \dots \wedge \mathbf{v}_d,$

which contains the vector $\mathbf{v}_i$ twice, and hence each term in the sum is the zero tensor.

Conversely, suppose $\mathbf{v}_1,\dots,\mathbf{v}_d \in \mathbf{V}$ are vectors such that

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d =\mathbf{0}.$

We must prove that $\{\mathbf{v}_1,\dots,\mathbf{v}_d\}$ is a linearly dependent set in $\mathbf{V}$. We will prove the (equivalent) contrapositive statement: if $\{\mathbf{v}_1,\dots,\mathbf{v}_d\}$ is a linearly independent set in $\mathbf{V},$ then

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d \neq \mathbf{0}.$

We prove this by induction on $\mathbf{d}.$ In the case $d=1,$ we have that $\{\mathbf{v}_1\}$ is linearly independent, so

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d = \mathbf{v}_d \neq \mathbf{0}.$

For the inductive step, we proceed as follows. Since $\{\mathbf{v}_1,\dots,\mathbf{v}_d\}$ is a linearly independent set, it is a basis of the subspace

$\mathbf{W} = \mathrm{Span}\{\mathbf{v}_1,\dots,\mathbf{v}_d\}$.

Let $\langle \cdot,\cdot \rangle$ denote the scalar product on $\mathbf{W}$ defined by declaring this basis to be orthonormal. We now define a linear transformation

$L \colon \mathbf{W}^{\wedge d} \to \mathbf{W}^{\wedge d-1}$

by

$L\mathbf{w}_1 \wedge \dots \wedge \mathbf{w}_d = \langle \mathbf{v}_1,\mathbf{w}_1\rangle \mathbf{w}_2 \wedge \dots \wedge \mathbf{w}_d.$

We then have that

$L\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d = \langle \mathbf{v}_1,\mathbf{v}_1\rangle \mathbf{v}_2 \wedge \dots \wedge \mathbf{v}_d = \mathbf{v}_2 \wedge \dots \wedge \mathbf{v}_d.$

Now, since $\{\mathbf{v}_1,\dots,\mathbf{v}_d\}$ is a linearly independent set, so is the subset $\{\mathbf{v}_2,\dots,\mathbf{v}_d\}.$ Thus, by the induction hypothesis,

$\mathbf{v}_2 \wedge \dots \mathbf{v}_d \neq 0.$

It then follows that

$\mathbf{v}_1 \wedge \dots \mathbf{v}_d \neq \mathbf{0},$

since otherwise the linear transformation $L$ would map the zero vector in $\mathbf{W}^{\wedge d}$ to a nonzero vector in $\mathbf{W}^{\wedge d-1},$ which is impossible.

— Q.E.D.

Corollary 1: We have $\|\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d\| \geq 0$ with equality if and only if $\{\mathbf{v}_1,\dots,\mathbf{v}_d\}$ is linearly dependent.

Since

$\|\|\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d\|^2 = \sum\limits_{\pi \in \mathrm{S}(d)} \mathrm{sgn}(\pi) \langle \mathbf{v}_1,\mathbf{v}_{\pi(1)}\rangle \dots \langle \mathbf{v}_d,\mathbf{v}_{\pi(d)}\rangle$

you can think of this as a massive generalization of the Cauchy-Schwarz inequality, which is the case $d=2.$

1. Finn says:

There is a typo in the definition of anti-symmetric tensor.

1. I’m not seeing it – a little help?

1. Thomas Allen says:

I believe there should be a wedge instead of a vee.

2. Isaiah says:

I may be wrong, but I believe when introducing the notation for Fun(d, N) that “N” should be lowercase and not uppercase. At least based on inference, it seems that n = N is the only way that section makes sense (where lowercase-n is the cardinality of the set of vectors {e_1, …, e_n} which make up an orthonormal basis of V).

Also, if I’m correct, just above that, it should say: “such a basis is simply given by all tensor products with d factors such that each factor is a vector in **E**” (E not V).