Contraction of Tensors is Independent of the Choice of Basis

Question

Definition: Let $ T:(V^*)^k \times V^l \rightarrow \mathbb{R}$ be a tensor of type $ (k,l)$. Let $ \{v_1,...,v_n\}$ be a basis of $V$ and $ \{v^{1^*},...,v^{n^*}\}$ be the corresponding dual basis. The contraction of $ T$ with respect to the $ i$th (dual vector) and $j$th slot (which is a tensor of type $(k-1,l-1)$) is given by

$\displaystyle CT=\sum_{k=1}^n T(\bullet,...,v^{k^*},...,\bullet; \bullet,...,v_k,...,\bullet)$

where the substitutions are done at the $i$th dual vector slot and the $j$th vector slot.

Theorem: The contraction of a tensor is well-defined, i.e. it is independent of the choice of the basis of $V$.

Proof: Recall that if $v \in V$, $w \in V^*$, then we have

$$\displaystyle v=\sum_{k=1}^n v^{k^*}(v)v_k$$

$$\displaystyle w=\sum_{k=1}^n w(v_k)v^{k^*}$$

Let $\{v_1',...,v_n'\}$ be another basis of $V$ and $\{v^{{1^*}'},...,v^{{n^*}'}\}$ be the corresponding dual basis. Then we have

$$\displaystyle v_i'=\sum_{k=1}^n v^{k^*}(v_i')v_k$$

$$\displaystyle v^{{i^*}'}=\sum_{k=1}^nv^{{i^*}'}(v_k)v^{k^*}$$

Hence, we get

$$\begin{aligned} \displaystyle \sum_{k=1}^n T(\bullet,...,v^{{k^*}'},...,\bullet; \bullet,...,v_k',...,\bullet) &=\sum_{k=1}^n T(\bullet,...,\sum_{j=1}^nv^{{k^*}'}(v_j)v^{j^*},...,\bullet; \bullet,...,\sum_{p=1}^n v^{p^*}(v_k')v_p,...,\bullet)\\&= \sum_{k=1}^n\sum_{p=1}^n\sum_{j=1}^n v^{{k^*}'}(v_j)v^{p^*}(v_k')T(\bullet,...,v^{j^*},...,\bullet; \bullet,...,v_p,...,\bullet)\\&=\sum_{k=1}^n\sum_{p=1}^n\sum_{j=1}^n v^{p^*}(v^{{k^*}'}(v_j)v_k')T(\bullet,...,v^{j^*},...,\bullet; \bullet,...,v_p,...,\bullet)\end{aligned}$$

I have no idea how to proceed. Can anyone give me some suggestions? Thanks.

Answer 1

You already wrote everything. In the last line you wrote $v^{k^{*'}}(v_j)v'_k$ but that is exactly $v_j$ as you wrote in the above relations. You substitute it in your equation and get $v^{p^*}(v_j)=\delta^{p^*}_j$ and conclude.

Answer 2

I don't have enough reputation to write a comment, so I'll write a short answer. You seem to make the problem rather complicated.

Let's simplify to the case of a (1,1) tensor $ \omega \otimes v$. Denote contraction by C, then $$C (\omega \otimes v )= \omega (v)=\omega ^i v_i.$$ Under a coordinate transformation, i.e. a change of basis, the components of $\omega $ and $v$ transform inverse to each other, hence contraction is independent of the basis.

Another way to see this is that a (1,1) tensor is a linear map $V\rightarrow V$ and contraction is the trace of this linear map.