Tutorial 3: Multilinear Algebra (International Winter School on Gravity and Light 2015)

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Recommends Ben H · 2022-12 · Mathematics Stack Exchange · on “What is the operation that happens between a tensor product and a vector under matrix multiplication?”

See, for example this section of Schuller's GR tutorials [this video] , where a $(1,1)$-tensor is used to create a map from a co-vector to a co-vector (instead of a map from a vector to a vector). And, of course, watch the associated video on Multilinear Algebra [LINK] by Schuller.

Cites drlang · 2017-05 · Mathematics Stack Exchange · on “Tensor rank explanation please.”

What is the difference between (2,0) and (1,1) and (0,2) tensors? The problem I was dealing with was a map $V \times V \to \Bbb R$ Would it be correct that a map V* x V would be (1,1) and a V* x V* would be a (2,0)? The question is regarding a tutorial ( [this video] ) at where at 29:54 he says the tensor is a (0,2).

Cites Taylor Rendon · 2023-08 · Mathematics Stack Exchange · on “Proving that $A^{[ab]}B_{ab} = A^{ab}B_{[ab]}$”

On this [this video] problem sheet, I am trying to show that for tensors (given a vector space V and it's dual $V^{*}$ ) $$A:V^{*} \times V^{*} \to \mathbb{R} \\ B: V \times V \to \mathbb{R}$$ that $A^{[ab]}B_{ab} = A^{ab}B_{[ab]}$ where $A^{[ab]} := \frac{1}{2}(A^{ab} - A^{ba})$ . This seems like a straightforward calculation, but I'm not understanding the following $(*)$ line in the solution: \begin{align*} A^{[ab]}B_{ab} &= \frac{1}{2}(A^{ab}B_{ab}-\color{red}{A^{ba}B_{ab}}) \\ &= \frac{1}{2}(A^{ab}B_{ab}-A^{mn}B_{nm}) \tag{$*$} \\ &= \frac{1}{2}(A^{ab}B_{ab}-A^{ab}B_{ba}) \\ &= A^{ab}B_{[ab]} \end{align*} The instructor insists that it doesn't matter how we name the indices being summed over,