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Given a constant symmetric tensor $V_{ij}$, calculate $V_{ij}V^{jk}$.

I don't see clearly how to do this. Is it possible to contract the $j$ index? How can I visualize this as a matrix multiplication?

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2 Answers 2

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It is a contraction of tensors. Use: $$ V_i^k=V_{ij}V^{jk}=\sum_{j=1}^nV_{ij}V^{jk}=V_{i1}V^{1k}+V_{i2}V^{2k}+...+V_{in}V^{nk} $$

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I assume that everything is over a vector space in coordinates wrt. a given basis, $V_{ij}$ represents a bilinear form $V$ and $V^{j,k}$ its metric transpose $V^*$, that is, a 2-form that to two $1$-forms $x^*:=g(x,\cdot)$ and $y^*:=g(y,\cdot) $assigns $V^*(x^*,y^*):=V(x,y)$, and further that $V_{ij} V^{jk}$ is the composition of tensor product and contraction...

This looks complicated, but you only need to sum over $j$ and obtain an entity $$\mathrm{Result}\,_i^{\,\,k}:=\sum_j V_{ij} V^{jk}.$$

Equivalently, you can compute it as $\sum_{j,u,v} V_{ij} g^{ju} V_{uv} g^{vk}=\sum_u V_i^{\,\,u} V_{u}^{\,\,k}$ hence if you represent $V$ as a "matrix" (=$(1,1)$-tensor) $V_x^{\,\,y}:=\sum_j V_{xj} g^{jy}$, then the result is just the square of this matrix.

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