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I have a polynomial function $f(x_1,x_2,x_3,x_4): \mathbb{C}^4 \to \mathbb{C}$, which obeys the equality $f(x_1+tx_3,x_2+tx_4,x_3,x_4) = f(x_1,x_2,x_3,x_4)$ for all $t \in \mathbb{C}$.

My question is essentially what can we say about $f$? More specifically, I suspect we can specify more about $f$ from this property, such as that (for example) it is actually only a polynomial in $x_3$ and $x_4$: all other terms with nonzero powers of $x_1$ and $x_2$ have zero coefficients. (I do not know if that's true, it's just an example of the sort of answer I'm looking for.)

Now I did originally think that my above example might be true, but I'm unable to formally prove it so I'm less convinced now. In particular, the fact that we translate our first two terms by $tx_3$ and $tx_4$ (rather than just $t$, say) complicates things.

I tried writing

$f(x_1,x_2,x_3,x_4) = \sum \limits_{i,j,k,l \geq 0} a_{ijkl}\, x_1^i x_2^j x_3^k x_4^l $

for some coefficients $a_{ijkl}$ (and with the sum consisting of finitely many terms); then

$\sum \limits_{i,j,k,l \geq 0} a_{ijkl}\, x_1^i x_2^j x_3^k x_4^l = \sum \limits_{i,j,k,l \geq 0} a_{ijkl}\, (x_1+tx_3)^i (x_2+tx_4)^j x_3^k x_4^l$ for all $t,x_1,x_2,x_3,x_4.\,\,\,\,(*)$

I then tried looking at this as either a polynomial in $t$ or a polynomial in $t,x_1,\ldots,x_4$; we know that all coefficients of $t^r$ are always zero for $r>0$, and these coefficients are polynomials in the $x_m$ and the $a_{ijkl}$; what I really want to do is show that these coefficients are necessarily nonzero polynomials in the $x_m$, unless we assume all $a_{ijkl}$ involved are zero.

However, when I try to determine the coefficient of $t^r$ as a poly in the $x_m$, I find that multiple terms in $(*)$ can contribute to the same term $x_1^{i'}x_2^{j'}x_3^{k'}x_4^{l'}$ in the coefficient of $t^r$, and so it isn't obvious to me that all the $a_{ijkl}$ involved in this coefficient should therefore be zero. (Sorry, I am aware that's a bit confusing to understand, I couldn't find a better way to phrase it.)

So, what can we say about this function? I had hoped to show $f(x_1,x_2,x_3,x_4)=h(x_3,x_4)$, but I haven't been able to. Indeed, functions such as these are included in our class of possible functions, but they may not make up the whole thing. Can we say something else about this class? Is there some particular form of polynomial which obeys this $t$-translation invariance?

(Edit: There may be a far nicer way to approach this problem than I've taken; this method is pretty ugly. If there is some nicer method please go ahead and use that rather than doing what I tried to do; the neater the better.)

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1 Answer 1

The function $x_1 x_4 - x_2 x_3$ satisfies your functional equation.

You can differentiate your relation with respect to $t$ to get $\frac{\partial f}{\partial x_1} x_3 + \frac{\partial f}{\partial x_2} x_4 = 0$.

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And this consists of -all- polynomials satisfying that relation, correct? –  T. Hughes Apr 27 '12 at 3:43
    
More generally, $f(x_1,x_2,x_3,x_4) = g(x_3,x_4,x_1 x_4 - x_2 x_3)$ for any polynomial $g$ in $3$ variables. –  Robert Israel Apr 27 '12 at 3:57
    
@RobertIsrael: How do you prove this explicitly, sorry? –  T. Hughes Apr 27 '12 at 5:26
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Ah, i guess it suffices to simply take $t=-x_1/x_3$. –  T. Hughes Apr 27 '12 at 5:34
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