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Is there a way to compute the following nullity of multilinear maps? As it is different from any nullity I know of, I call it bunnity after myself:-)) If it already has a name, it be nice to know it. References are particularly appreciated.

Let $V$ be a $d$-dimensional vector space over a field $K$, $f:V^{\otimes n}\rightarrow K$ a multilinear map. We define the bunnity by

$Bun(f) = sup \{ \sum_{k=1}^n dim(U_k) \ | \ f(U_1\otimes U_2 \ldots \otimes U_n)=0\}$.

If $n=2$, the bunnity is nullity plus dimension. It is more complicated if $n\geq 3$. I expect that for $n=3$ or $n=4$, there should be an algorithm because the 3-subspace problem is finite, while the 4-subspace problem is tame. I would be pleasantly surprised if there is an algorithm for a general $n$.

EDIT I forgot to say what $U_i$ are. They are non-zero vector subspaces of $V$.

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  • $\begingroup$ What's the connection to multilinear tensor rank? $\endgroup$
    – Suvrit
    Commented May 10, 2015 at 14:39
  • $\begingroup$ Inequality. In the usual tensor rank/nullity, you choose each $U_i$ separately. I ask you to choose them simultaneously. $\endgroup$
    – Bugs Bunny
    Commented May 10, 2015 at 14:42
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    $\begingroup$ You should offer a bounty to whoever manages to frame the answer in terms of Wascal's triangle. $\endgroup$
    – Vidit Nanda
    Commented May 10, 2015 at 16:19
  • $\begingroup$ Could you remind what definition of "nullity" do you use (e.g. when $n=2$)? $\endgroup$
    – YCor
    Commented May 14, 2015 at 12:48
  • $\begingroup$ The dimension of the null-space: think of $f$ as a map $V\rightarrow V^\ast$ and take the dimension of its kernel. Similarly, the usual multinullity is defined for any $n$: use position $i$ to write $f$ as $V\rightarrow (V^{\otimes (n-1)})^\ast$ and take the dimension of its kernel. $\endgroup$
    – Bugs Bunny
    Commented May 15, 2015 at 8:59

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