# Identifying $k[x_1,x_2,y_1,y_2]^{\epsilon}$ with $k[x,y]\wedge k[x,y]$

Suppose the symmetric group $S_2$ of order 2 acts on $k^4=Spec \;k[x_1, x_2, y_1, y_2]$ by the following: for $\sigma\not=e$, $$\sigma\circ(x_1, x_2, y_1, y_2)=(x_2,x_1,y_2,y_1).$$

That is, the nontrivial element in $S_2$ swaps the indeterminants $x_1$ and $x_2$, and $y_1$ and $y_2$, simultaneously.

Let $k[x_1, x_2,y_1, y_2]^{\epsilon}$ denote the vector subspace of $k[x_1, x_2,y_1, y_2]$ consisting of $S_2$-alternating polynomials, and let $k[x,y]\wedge k[x,y]$ denote the $2^{nd}$ exterior power of the vector space $k[x,y]$ of polynomials in $2$ variables.

Then how is $k[x_1, x_2,y_1, y_2]^{\epsilon}$ identified with $k[x,y]\wedge k[x,y]$?

$\mathbf{Examples}:$ Some examples of polynomials in $k[x_1, x_2,y_1, y_2]^{\epsilon}$ include the following: $$g_1 = x_1-x_2,\; g_2= x_1-x_2+y_1-y_2, \;\mbox{ and } \;g_3 = (x_1 + x_2)(x_1-x_2)$$

since $g_i(x_2,x_1,y_2,y_1)=-g_i(x_1,x_2,y_1,y_2)$. This basically means when you swap the $x_i$'s and the $y_i$'s, the polynomial only changes by a sign.

At the moment, this identification is not obvious to me.



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Hints:

• We have an isomorphism $A:=k[x_1,x_2,y_1,y_2]\cong k[x_1,y_1]\otimes_k k[x_2,y_2]=B$.
• The action $\tau:A\to A:p\mapsto -\sigma p$ is a vector space homomorphism.
• The action of $\tau$ on $B$ is generated by $p(x_1,y_1)\otimes q(x_2,y_2)\mapsto -q(x_1,y_1)\otimes p(x_2,y_2)$.
• If $\varphi\in \mathrm{End}(V)$ is a linear map, $V/\mathrm{Im}(I-\varphi)\cong \mathrm{Ker}(I-\varphi)= V_1$, where $V_1$ is the eigenspace corresponding to the eigenvalue $1$ (i.e. the fixed points of the transformation).

(I'm going on the definition that $\bigwedge^2 V:= V\otimes V/I$, where $I$ is the subspace spanned by those elements of the form $v\otimes w+w\otimes v$ with $v,w\in V$.)

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Thanks for the hints anon, but I am not sure how the third hint comes in. Do you mean $I$ to be the identity matrix? – math-visitor Jun 16 '12 at 8:29
@math-visitor: Yes. If $\varphi$ is the nontrivial $S_2$ action (well, its negation), what do the image and kernel of $I-\varphi$ look like on the tensor product given in the first bullet point? – anon Jun 16 '12 at 8:30
There was an error in my original answer and I couldn't figure out what it was until just now. We need $k[x_1,y_1]\otimes k[x_2,y_2]$ rather than $k[x_1,x_2]\otimes k[y_1,y_2]$ for this to work. – anon Jun 16 '12 at 9:00
I am wondering that in the second line, if $p$ is supposed to map to $\sigma p$, not $-\sigma p$. In the third line, shouldn't $p(x_1,y_1)\otimes q(x_2,y_2)$ map to $-p(x_2,y_2)\otimes q(x_1,y_1)$... – math-visitor Jun 16 '12 at 9:23
No and no. :) We want $p\mapsto -\sigma p$, because then the fixed points of this action are the $S_2$-alternating polynomials (when speaking on $A$). On $B$, it suffices to check where $\tau$ sends the pure tensors $p(x_1,y_1)\otimes q(x_2,y_2)$, which correspond to the separable polynomials $p(x_1,y_1)q(x_2,y_2)$ in $A$ ... – anon Jun 16 '12 at 9:26