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Jul 23 at 10:54 comment added Antoine @DaveBenson I agree!
Jul 18 at 6:49 comment added Dave Benson @Antoine I think he must mean $f(w_i,v_j)=\delta_{i,j}$.
Jul 18 at 4:31 comment added Antoine Well, the thing is that your formula for $f(v,v')$ implies that $f(v_i,w_j)=0$ for all $i,j$.
Jul 17 at 14:55 comment added testaccount With what I wrote $vv' = f(v,v') \prod_k v_k^{i_k+i_k'} \prod_k w_k^{j_k+j_k'}$ in $P$, isn't that correct?
Jul 17 at 5:50 comment added Antoine thank you, I guess you mean $f(v,v')=z^{\sum_{k=1}^n{i_k j'_k}}$, right?
Jul 17 at 2:57 history answered testaccount CC BY-SA 4.0