# Showing that a function is a tensor

I am trying to solve question 4 in Munkres Analysis on Manifolds section 26.

The question is determine if the following is a tensor on $\mathbb{R}^4$ and express those that are in terms of the elementary tensors on $\mathbb{R}^4$.

$f(x,y) = 3 x_1 y_2 + 5 x_2 x_3$

I am trying to follow the definition of a tensor (which I don't think I fully understand) to show this. I know I must find a function $T: \mathbb{R}^4 \rightarrow \mathbb{R}$ that is linear on the $i^{th}$ variable. Since there are only 2, then I believe I have to somehow write $f$ as a projection. I'm just not sure where to go from here.

Any hints would be great, thanks!

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Perhaps I am missing something, but you have $f(\lambda x,0) = \lambda^2 f(x,0)$. Hence it is not linear in $x$, hence not a tensor.

If $f$ was a tensor, it would have the form $f:\mathbb{R}^4 \times \mathbb{R}^4 \to \mathbb{R}$, and be linear in each variable separately, that is the functions $w_1 \mapsto f(w_1,y)$ and $w_2 \mapsto f(x,w_2)$ would be linear for all $x,y$.

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I see now, the definition is much more clear. Thank you for your help. This should be sufficient for me to move on with the chapter. – Ezea Nov 25 '12 at 0:10
I have verified that $h(x,y) = x_1 y_1 - 7x_2 y_3$ is a tensor and am asked to represent it in terms of the elementary tensors on $\mathbb{R}^4$. I am slightly confused about how to do this. I know somehow I want to write sums that multiply by the basis to give me $x_1 y_1$ with ever other term 0 and $-7x_2 y_3$ with every other term 0. Am I on the right track? – Ezea Nov 25 '12 at 19:49
Let $e_i^*$ be the functional on $\mathbb{R}^4$ defined by $e_i^*(e_k) = \delta_{ik}$ (ie, the dual basis). Given functionals $f,g$ on $\mathbb{R}^4$, define the tensor product $(f \otimes g) (x,y) = f(x)g(y)$. You can show that the elementary tensors $e_j^* \otimes e_k^*$ form a basis for the $2$-tensors on $\mathbb{R}^4$. Note that the tensor $(x,y) \mapsto -7 x_2 y_3$ can be written as $(x,y) \mapsto -7 (e_2^* \otimes e_3^*)(x,y)$. Hence you have $h = e_1^* \otimes e_1^* - 7 e_2^* \otimes e_3^*$. – copper.hat Nov 25 '12 at 23:34

edit: Oops, I misread it as $f(x, y) = 3x_1y_2 + 5x_2y_3$ . But still, the answer serves as an example of how to verify if it's a tensor.

Just compute $f(x+z, y) = 3(x_1+z_1)y_2 + 5(x_2+z_2)y_3 = 3x_1y_2 + 3z_1y_2 + 5x_2y_3 + 5z_2y_3 = f(x, y) + f(z, y)$
Repeat for $f(x, y+z)$, $f(cx, y)$ and $f(x, cy)$to verify that it's a tensor.

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I think you have used $y_3$ instead of $x_3$? – copper.hat Nov 25 '12 at 0:04
Ok, thank you. I can compute these and I see intuitively why that would mean it is linear for those variables. I will have to try to interpret the definition like this. I did not see that as my strategy from what I've seen so far. – Ezea Nov 25 '12 at 0:04