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After much shifting through notational hurdles, I may have gotten the point. However, I'd like to confirm this unequivocally by working through an example.

If $\beta \in V^*$ is $\beta=\begin{bmatrix}1 &2 &3 \end{bmatrix}$ and $\gamma\in V^*$ is $\gamma=\begin{bmatrix}2 &4 &6 \end{bmatrix}$. The $(2,0)$-tensor $\beta\otimes \gamma$ is the outer product:

$$\beta\otimes_o \gamma=\begin{bmatrix}2\,e^1\otimes e^1&4\,e^1\otimes e^2&6\,e^1\otimes e^3\\4\,e^2\otimes e^1&8\,e^2\otimes e^2&12\,e^2\otimes e^3\\6\,e^3\otimes e^1&12\,e^3\otimes e^2&18\,e^3\otimes e^3\end{bmatrix}$$

Now if apply this tensor product on the vectors

$$v=\begin{bmatrix}1\\-1\\5\end{bmatrix}, \; w = \begin{bmatrix}2\\0\\3\end{bmatrix}$$

$$\begin{align} (\beta \otimes \gamma)[v,w]&=\\[2ex] & 2 \times 1 \times 2 \quad+\quad 4 \times 1 \times 0 \quad +\quad 6 \times 1 \times 3 \\ +\;&4 \times -1 \times 2 \quad + \quad 8 \times -1 \times 0 \quad + \quad 12 \times -1 \times 3 \\ +\;&6 \times 5 \times 2 \quad + \quad 12 \times 5 \times 0 \quad + \quad 18 \times 5 \times 3 \\[2ex] &= 308\end{align}$$

Is this correct?

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  • $\begingroup$ Yes, this is correct. The answer should be $\left<\beta,v\right> \cdot \left<\gamma,w\right>$, where $\left<,\right>$ is the usual inner product. $\endgroup$ Commented Sep 20, 2017 at 15:15
  • $\begingroup$ @Nick Thank you. And indeed... $\vec \beta \cdot \vec v \times \vec \gamma \cdot \vec w = 308.$ v = c(1,-1,5); w = c(2,0,3); beta = 1:3; gamma = c(2,4,6); beta %*% v * gamma %*% w 308. $\endgroup$ Commented Sep 20, 2017 at 15:30
  • $\begingroup$ @Nick Is it always that easy? I mean, when you start adding additional covectors and vectors to the tensor product... $\endgroup$ Commented Sep 20, 2017 at 15:32

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Here is a further explanation of my comment above:

For linear maps $f \colon V \to X$ and $g \colon W \to Y$, the map $f \otimes g \colon V \otimes W \to X \otimes Y$ is defined by

$$ (f \otimes g)(v \otimes w) = f(v) \otimes g(w) $$

In your case, $\beta$ and $\gamma$ are both elements of $V^* = \mathrm{Hom}(V,\mathbb{K})$, and so $\beta \otimes \gamma$ is a linear map $V \otimes V \to \Bbb{K} \otimes \Bbb{K}$, given by

$$ (\beta \otimes \gamma)(v \otimes w) = \beta(v) \otimes \gamma(w) $$

Now, $\Bbb{K} \otimes \Bbb{K} \cong \Bbb{K}$ by the identification $a \otimes b \mapsto a \cdot b$. So this is how we think of $\beta \otimes \gamma$ as a bilinear map given by:

$$ (\beta \otimes \gamma)(v,w) = \beta(v) \cdot \gamma(w) $$

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  • $\begingroup$ Thank you. Much more abstract... There is (I presume) a tiny $LaTeX$ issue in the first equation, where it should be $(f\otimes g)(v\otimes w)$(?) Also, I wonder if you could explain more basically the idea in the last paragraph, and make a connection of the post with the concrete calculation in your comment, $\left<\beta,v\right> \cdot \left<\gamma,w\right>$ as it applies to tensors in general. $\endgroup$ Commented Sep 20, 2017 at 15:43
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    $\begingroup$ If $\beta \in V^*$, then $\beta(v)$ and $\left<\beta,v\right>$ mean the same thing, depending on if you'd rather think of $\beta$ as a function or a vector. $\endgroup$ Commented Sep 20, 2017 at 15:45

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