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I'm trying to decide which of the following are alternating tensors in $\mathbb R^4$ and express those that are in terms of the elementary tensors on R^4:

$f(x,y) = x_1y_2 = x_2y_1 + x_1y_1$ \begin{align*} g(x, y) &= x_1y_3 - x_3y_2 \\ h(x, y) &= x_1^3y_2^3 - x_2^3y_1^3 \end{align*}

All I can tell is that we have to define a $\sigma$ permutation by $f(\sigma(x),\sigma(y) =$

and I'm trying to go about solving it.

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I tried to clean this up, but the formula for $f$ doesn't make sense as written and the remarks after the display don't seem to be finished. Please edit! – Dylan Moreland Mar 19 '12 at 2:00
It looks like your "tensors" are maps $(\mathbf R^4)^2 \to \mathbf R$ and are functions of two arguments. In that case, if you have shown that one of them is linear in each argument ($h$ looks somewhat suspect), you really only need to check that $f(y, x) = -f(x, y)$ for all $x, y \in \mathbf R^4$. This is just the condition for being an alternating bilinear form, so there's no need to think too hard about permutations. – Dylan Moreland Mar 19 '12 at 2:02
If the second $=$ in the line describing $f$ means to be an $=$, or something else? And the sentence "All I can tell is that we have to define..." is incomplete (and the formula has unbalanced parentheses, not to mention a completely AWOL right hand side of the equality!) – Arturo Magidin Mar 19 '12 at 2:18
I will edit it above – mary Mar 19 '12 at 6:02
Thanks for catching the incompleteness – mary Mar 19 '12 at 6:04

If I'm interpreting your symbols correctly: you want to check whether these maps $\mathbf R^4 \times \mathbf R^4 \to \mathbf R$ are alternating bilinear forms. All such maps have the following form: let $G$ be a $4 \times 4$ matrix which is skew-symmetric (that is, ${}^tG = -G$), and define a form by \[ (x, y) \mapsto {}^txGy. \] If we forget about the alternating ($\Leftrightarrow$ $G$ skew-symmetric) condition for a moment, $h$ corresponds to the matrix \[ G = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & -1 & 0 \end{pmatrix}. \] What conclusions can you draw? On the other hand, $h$ doesn't look like something that you can write in this form at all. For example, if $x = (1, 0, 0, 0)$ and $y = (0, 1, 0, 0)$ we ought to have $h(2x, y) = 2h(x, y)$. Do we?

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Mary, would you care to comment on this answer? – The Chaz 2.0 Mar 20 '12 at 13:05

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