math121b:01-27

# 2020-01-27, Monday

Last time, we have reviewed the abstract definition of vector space. And we defined the tensor products of any two vector spaces $V$ and $W$, as a new vector space $V \otimes W$. Today, we introduce the notion of the dual vector space $V^*$ of a given vector space $V$. Then, if $V$ is equipped with an inner product, we construc the metric tensor $g \in V^* \otimes V^*$ which express the same information.

Let $V$ be a finite dimensional vector space over $\R$ throughout this note.

## Dual Vector Space

We let $V^*$ denote the set of linear functions on $V$. One can verify that $V^*$ is also a vector space over $\R$. If $\dim V=n$, then $\dim V^*=n$ as well.

Dual basis Let $e_1, \cdots, e_n$ be a basis of $V$, to specify an element in $V^*$, we just need to specify its value on the basis elements. We define the following elements $h_1, \cdots, h_n$ in $V^*$: $$h_i (e_j) = \delta_{ij}$$ One can show that $h_i$ forms a basis of $V^*$. $\{h_i\}$ is said to be the dual basis of $\{e_i\}$.

Canonical Pairing There is a canonical pairing between $V$ and $V^*$, denoted as $$\langle -, -\rangle: V \times V^* \to \R, \quad (v, h) \mapsto h(v)$$.

$$\gdef\ot\otimes$$

More generally, we extend the pairing on tensor products $V^* \otimes V^*$ and $V \otimes V$ $$\langle -, -\rangle: (V \otimes V) \times (V^* \otimes V^*) \to \R$$ where $$\langle v_1 \ot v_2, h_1 \ot h_2 \rangle = h_1(v_1) h_2(v_2).$$ In fact, we have $V^* \ot V^* \cong (V \ot V)^*$.

## Inner product and metric tensor

Recall that an inner product on $V$ is a positive definite symmetric pairing on $V$ $$(-, -): V \times V \to \R$$ where if $v \in V$ is non-zero, we have $(v, v) > 0$. We define $\| v\|^2 = (v,v)$.

Let $V$ be a finite dimensional vector space with inner product (a.k.a Euclidean vector space). The metric tensor $g$ of $V$, is a tensor $g \in V^* \otimes V^*$, defined uniquely by the following requirement $$\langle g, v \ot w \rangle = (v,w).$$

If $e_1,\cdots, \e_n$ are a ortho-normal basis of $V$, and $h_1, \cdots, h_n$ are the dual basis. Then we may write $g$ as $$g = \sum_{i=1}^n h_i \otimes h_i.$$

In general, for any basis $e_1, \cdots, e_n$ and corresponding dual basis $h_1, \cdots, h_n$, we have $$g = \sum_{i,j=1}^n (e_i, e_j) h_i \otimes h_j$$

math121b/01-27.txt · Last modified: 2020/01/27 13:00 (external edit)