1
$\begingroup$

I am looking for a direct proof (i.e. without going through ABPs) that if $f(\bar{x})$ has an arithmetic formula of size $s$ then it is a projection of an $O(s)\times O(s)$ determinant.

It seems possible using induction on $s$.

The basis, when $s=1,2,3$ is trivial.

Suppose we proved it for size $s$ formulas and let $f$ be a size $s+1$ formula.

If the root gate is a $\times$ gate with children $f_1,\cdots,f_k$, by induction we have that each $f_i$, which is also a formula, is a projection of a determinant of size $O(s_i)\times O(s_i)$, say $f_i=\text{det}(M_i)$.

So, if we construct a new matrix $M$ which is a block matrix with the $M_i$'s as its blocks and zeros elsewhere. Then $\text{det}(M)=\prod_{i=1}^k\text{det}(M_i)$ and the dimension of $M$ is $O(s_1)+\cdots+O(s_k)=O(s)$.

My problem is when the root is a $+$ gate.

$\endgroup$
3
  • $\begingroup$ You can find this proof in Valiant’s original paper, Completeness classes in algebra. $\endgroup$ – Yuval Filmus Jun 1 '18 at 21:37
  • $\begingroup$ @YuvalFilmus What he does in the paper is going through ABPs without naming them. $\endgroup$ – Don Fanucci Jun 1 '18 at 21:48
  • $\begingroup$ There should be a proof on the monograph Algebraic Complexity Theory. Avi Wigderson explained the proof in a lecture once, so it definitely exists, and isn't too difficult; though a bit tricky. The key to the induction is that the matrix has to be of a particular form. $\endgroup$ – Yuval Filmus Jun 1 '18 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.