Tensors with low spectral norm

2012-05-16T20:03:03Z

Consider a tensor $T$ with six indices, $T_{(ii')(jj')(kk')}$, where each index goes from $1$ to $n$. We can think of $T$ as a linear map from $\mathbb{R}^n \otimes \mathbb{R}^n \otimes \mathbb{R}^n$ to itself and consider its spectral norm:

$ \Vert T \Vert_{\infty} = \sup \vert \sum T_{(ii')(jj')(kk')} x_{ijk}y_{i'j'k'}\vert $

where the supremum is taken over all unit vectors $x,y \in \mathbb{R}^n \otimes \mathbb{R}^n \otimes \mathbb{R}^n$.

On the other hand $T$ can also be viewed as trilinear form on $n$-dimensional matrices, so we can define a norm:

$ \Vert T \Vert_{2,2,2} = \sup \vert \sum T_{(ii')(jj')(kk')} X_{ii'} Y_{jj'}Z_{kk'}\vert $

where the supremum is over all matrices $X,Y,Z$ of Frobenius norm $1$.

What are the examples of tensors which have high (as $n \to \infty$) spectral norm as linear maps, but low norm as trilinear forms?

Since the question is admittedly rather general, let's specialize to tensors of a more special form, namely $T_{(ii')(jj')(kk')} = g_{ijk}h_{i'j'k'}$, where $g,h \in \mathbb{R}^n \otimes \mathbb{R}^n \otimes \mathbb{R}^n$, so that $T = gh^{T}$ as a linear map. Then its spectral norm is simply $\Vert g\Vert \cdot \Vert h\Vert$. Taking $g$ and $h$, say, to be unit vectors, how should we choose them to get a low trilinear norm:

$\Vert T \Vert_{2,2,2} = \sup \vert \sum g_{ijk}h_{i'j'k'} X_{ii'} Y_{jj'}Z_{kk'}\vert $ ?

It can be shown that randomly chosen $g, h$ will have the desired property, but I'm interested in more explicit examples. Because there is no direct analog of the spectral decomposition for tensors, the intuition that the "mass" of $T$ should be "spread out" roughly in all directions (as in the case of matrices with low spectral norm, but high Frobenius norm) on the $\mathbb{R}^{n^2}$ components of the tensor product is not easy to formalize.

Tensors with low spectral norm - MathOverflow

Tensors with low spectral norm