# Reduction of $f:\mathbb{C}^n\to \mathbb{C}$ into sum of $f_{ij}:\mathbb{C}^2\to\mathbb{C}$

I was browsing wikipedia the other day when I came across the following (paraphrased) claim: $$\exists f_{ij}:\mathbb{C}^2\to \mathbb{C} \mbox{ s.t. } f(x_1,\dots,x_n)=\sum_{i,j} f_{ij}(x_i,x_j)$$ Now for the life of me, I can't find that page and I'm not sure how I would search the literature for such a claim. I'm having trouble believing it unconditionally, but I would like to know under which conditions it is true.

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Is there an assumption about $f$? – Davide Giraudo May 24 '12 at 20:13
I would assume $f$ to be (sufficiently) smooth and square integrable. – Deathbreath May 24 '12 at 20:37
It seems to be not true, for example with $f(x_1,x_2,x_3)=x_1x_2x_3$. – Davide Giraudo May 24 '12 at 20:41
Can you give a link for this claim? – Davide Giraudo May 24 '12 at 21:30
@DavideGiraudo: The lack of a link is part of my problem ;-) – Deathbreath May 25 '12 at 13:28

For $n=3$, the existence of the desired decomposition of $f$ is equivalent to the equation $$f(x_1,x_2,x_3)+f(x_1,x_2',x_3')+f(x_1',x_2,x_3')+f(x_1',x_2',x_3)\\ =f(x_1,x_2,x_3')+f(x_1,x_2',z_3)+f(x_1',x_2,x_3)+f(x_1',x_2',x_3')$$ holding identically.

It can be directly checked that, if $f$ has a decomposition of the desired form, then the identity above holds. Conversely, if the identity above holds, then (setting $x_1'=x_2'=x_3'=0$): $$f(x_1,x_2,x_3)=f(x_1,x_2,0)+f(x_1,0,x_3)+f(0,x_2,x_3)-f(x_1,0,0)-f(0,x_2,0)-f(0,0,x_3)+f(0,0,0)$$ where we have expressed $f(x_1,x_2,x_3)$ as a sum of functions, each of which depends on at most two of the variables.

I would guess there are similar identities (or families of identities) for $n\geq 4$, but I'm not sure what they are.

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This is an ad hoc expansion of Pink Elephants' method, if anyone's still paying attention to this question.

Given $f: \mathbb{C}^n \to \mathbb{C}$, define the "Euler characteristic" as the function $\chi(f)$ in $2n$ variables given by $$\chi(f) = f(x_1, \ldots, x_n) - \sum_i f(x_1, \ldots, x_i',x_{i+1}, \ldots, x_n) + \sum_{i,j} f(x_1, \ldots, x_i', \ldots, x_j', \ldots, x_n) - \ldots + (-1)^nf(x_1', x_2', \ldots, x_n').$$

Then it looks like $f$ is a sum of functions from $\mathbb{C}^{n-1}$ if and only if $\chi(f) = 0$ (identically). You could iterate the definition of $\chi$ to get a criterion to reduce from $n$ to $2$.

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So it would be true if $\exists x_i^\prime: \chi(f)=0$, not just $\forall x_i^\prime$. Maybe I can use $\chi$ to develop an optimal approximation. – Deathbreath Jul 6 '12 at 13:43