math214:02-21

# 2020-02-21, Friday

$$\gdef\ot\otimes$$

## 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 $$e_{i_1} \wedge \cdots \wedge e_{i_k}, \quad 1 \leq i_1 < i_2 < \cdots < i_k \leq n.$$

Corrollary

• $\dim \wedge^k(V) = {n \choose k}$.
• 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 $$A \wedge B := v_1 \wedge \cdots \wedge v_k \wedge w_1 \wedge \cdots \wedge w_l \in \wedge^{k+1} V.$$

### Relationship between quotient algebra and tensor algebra

1. As quotient of tensor algebra
2. As subalgebra of tensor algebra

## Differential Forms

$\Omega^k(M) = \wedge^k T^*M.$

One can do integration of $k$-forms on $k$-submanifold.

## exterior differentiation

In local coordinates $(x_1, \cdots, x_n)$, given a differential $k$-form, we have $$d (\sum_I f_I dx^I) = \sum_I d(f_I) \wedge dx^I = \sum_I \sum_{i=1}^n \frac{\d f_I}{\d x^i} dx^i \wedge dx^I$$ 