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2020-01-29, Wednesday


Tensor power of a vector space

Let $V$ be a finite dimensional vector space.

We denote the $k$-th tensor power of $V$ as $$ V^{\otimes k} = \underbrace{V\ot \cdots \ot V}_{\text{ $k$ times} } $$ Its elements are linear combinations of terms like $v_1 \otimes \cdots \ot v_k$, subject to the usual linearity relations.

It is sometimes useful to consider the tensor algebra (we only mention it here, but do not use it later in this course).

Definition (Tensor Algebra $T(V)$ ) $$T(V) = \R \oplus V \oplus V^{\ot 2} \oplus \cdots \oplus V^{\ot 3} \oplus \cdots $$ Given two elements $T = w_1 \ot \cdots \ot w_k$ and $T' = v_1 \ot \cdots \ot v_l$, their products is defined by juxtapostion. $$ T \ot T' = w_1 \ot \cdots \ot w_k \ot v_1 \ot \cdots \ot v_l $$

Exterior power of a vector space

Definition (Exterior product $\wedge^k(V)$) The $k$-th exterior product $\wedge^k(V)$ is the vector space consisting of linear combinations of the following terms $v_1 \wedge \cdots \wedge v_k$, where the expression is linear in each slot, $$ c \cdot (v_1 \wedge \cdots \wedge v_k) = (c v_1) \wedge v_2 \wedge \cdots \wedge v_k $$ $$ (v_1+v_1') \wedge \cdots \wedge v_k = v_1 \wedge \cdots \wedge v_k + v_1'\wedge \cdots \wedge v_k$$ and the expression changes signs if we swap any two slots $$ v_1 \wedge \cdots \wedge v_i \wedge \cdots \wedge v_j\wedge \cdots \wedge v_k = - v_1 \wedge \cdots \wedge v_j \wedge \cdots \wedge v_i\wedge \cdots \wedge v_k, \forall 1 \leq i < j \leq k. $$

If $k=0$, we set $\wedge^0 V = \R$. If $k=1$, then $\wedge^1 V =V$.

Proposition If we choose a basis $e_1, \cdots, e_n$ of $V$, then for $0 \leq k \leq n$, the space $\wedge^k(V)$ has a basis consisting of the following vectors $$ e_{i_1} \wedge \cdots \wedge e_{i_k}, \quad 1 \leq i_1 < i_2 < \cdots < i_k \leq n. $$


  • $\dim \wedge^k(V) = {n \choose k}$.
  • If $k > n$, then $\wedge^k V = 0$.

Just as we defined tensor algebra $T(V)$, we may define the exterior algebra $\wedge^* V$. This turns out to be very useful.

Definition(Exterior algebra $\wedge^* V$) $$ \wedge^* V := \bigoplus_{k=0}^{n} \wedge^k V, \quad \text{ where } \wedge^0 V:= \R.$$ The product between two elements is given by juxtaposition, more precisely, if $A = v_1 \wedge \cdots \wedge v_k \in \wedge^k V$, $B = w_1 \wedge \cdots \wedge w_l \in \wedge^l V$, then $$ A \wedge B := v_1 \wedge \cdots \wedge v_k \wedge w_1 \wedge \cdots \wedge w_l \in \wedge^{k+1} V.$$

Example of $\R^3$

Let $V = \R^3$, and equip $V$ with the standard inner product.

Cross product

Consider the $\wedge^2 V$, its dimension is ${3 \choose 2} = 3$, hence element of it are sometimes called pseudo-tensor. If we use the standard basis $e_1, e_2, e_3$ on $V$, then we have the following basis for $\wedge^2 V$: $$ e_1 \wedge e_2, \quad e_1 \wedge e_3, \quad e_2 \wedge e_3. $$

There is a bijection from $\wedge^2 V \to V$, called “Hodge star” $\star$, which goes as follows: $$ \star: e_1 \wedge e_2 \mapsto e_3, \quad e_2 \wedge e_3 \mapsto e_1, \quad e_3 \wedge e_1 \mapsto e_2. $$

Thus, we may recover our familiar cross-product $\b v \times \b w$ formula as following $$ V \times V \xrightarrow{\wedge} \wedge^2 V \xrightarrow{\star} V. $$

Exercise: convince yourself that $\b v \wedge \b w = \star(\b v \wedge \b w)$.

Volume Forms and Determinant

For $\wedge^3 V$, it is one-dimensional, with $e_1 \wedge e_2 \wedge e_3$ as a basis. Thus, element of $\wedge^* V$ are sometimes called pseudo-scalar.

Given three vectors $v_1, v_2, v_3$, how to compute the signed volume formed by the parallelogram $P(v_1, v_2, v_3)$ (skewed boxes) with sides $v_1, v_2, v_3$?

From vector calculus, we know the answer is the determinant of the $3$ by $3$ matrix, whose column-vectors are $v_1, v_2, v_3$. $$ \text{ Volume of } P(v_1, v_2, v_3) = \det (v_1, v_2, v_3) = \det \begin{pmatrix} v_{11} & v_{21} & v_{31} \cr v_{12} & v_{22} & v_{32} \cr v_{13} & v_{23} & v_{33} \end{pmatrix} $$

Now, we have another way to express it. $$ \text{ Volume of } P(v_1, v_2, v_3) = \frac{ v_1 \wedge v_2 \wedge v_3}{e_1 \wedge e_2 \wedge e_3} $$ Indeed, since both the numerator and denominators are elements of the one-dim vector space $\wedge^3 V$, the raio makes sense.

Levi-Cevita Symbol and Kronecker Symbol

These are two symbols introduced in Boas's book. We list their definitions and some properties.

The Kronecker symbol is used everywhere $$ \delta_{ij} = \begin{cases} 1 & \text{if } i=j \cr 0 & \text{if } i \neq j \cr \end{cases} $$

The Levi-Cevita Symbol is a rank-3 tensor on $\R^3$. Its component with respect to the standard basis is $$ \epsilon_{ijk} = \begin{cases} 1 & \text{if } ijk=123,231,312 \cr -1 & \text{if } ijk = 213,132,321 \cr 0 & \text{ if $ijk$ has repeated indices } \end{cases} $$

For example, we can use $\epsilon_{ijk}$ to express the determinant $$\det(v_1, v_2, v_3) = \sum_{ijk} \epsilon_{ijk} v_{1i} v_{2j} v_{3k}. $$

The Levi-Cevita symbol has generalization to higher dimension. It is a rank n tensor on $\R^n$, i.e, it has $n$ indices. Let $I$ denote the index. $\epsilon_I=1$ if $I$ can be obtained from $12 \cdots n$ by even number of permutation, and $\epsilon_I=-1$ if $I$ can be similarly obtained by odd number of permutations, and $\epsilon_I=0$ if there are repeated indices in $I$.

Another useful property is that $$ \sum_i \epsilon_{ijk} \epsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm} \delta_{kl} $$

math121b/01-29.txt · Last modified: 2020/01/29 10:44 (external edit)