I've recently been interested in studying tensors and networks of tensors. Consider the following quantity: $$Tr(A^N)$$ Where $A$ is a square matrix and $N$ is large. This is extremely easy - all we need to do is find the eigenvalues and power each to $N$. In the limit that $N$ grows large, the trace is dominated by the largest eigenvalue. In tensor network notation, this is just a bunch of vertices on a line. The vertices are the matrices, and the lines connecting them represent contraction of the corresponding index.


My problem concerns the analogous scenario with a tensor $W_{abcd}$. Instead of contracting in a line, there are multiple copies of this tensor are arranged in an $N$ x $M$ square lattice and contracted with their nearest neighbors. The north, south, east, and west legs correspond to the a, b, c, and d indices. Is there a way to see how the 'eigenvalues' of this new tensor, or even the trace of the whole quantity, scale with $N$ and $M$? I'm pretty confident that the trace scales as the largest eigenvalue powered to the area of the rectangle, but I'm not sure how to prove it.

The first picture is $A^N$. The second is Tr($A^N$). The third is the trace of the $W$ matrices I want to consider.

EDIT: Also, even some simple info about related work on this topic would be greatly appreciated.

EDIT 2: Something of note is that if we don't trace out the uncontracted indices, the number of uncontracted indices will scale as the perimeter of the rectangle, and thus the tensor will grow in dimension of its indices.

EDIT 3: Some pictures!

  • $\begingroup$ So by "trace of the whole quantity" you mean to make the edges wrap around? Interesting question! $\endgroup$ – Oscar Cunningham Mar 16 '16 at 11:03
  • $\begingroup$ Yep, that's right. $\endgroup$ – Aurey Mar 17 '16 at 19:46
  • $\begingroup$ Given that you're talking in terms of graphical notation for the contraction of relevant indices, it seems like it'd be appropriate to include the relevant diagrams in your statement of the problem. $\endgroup$ – Semiclassical Mar 19 '16 at 1:14
  • $\begingroup$ You got it. Picture added. $\endgroup$ – Aurey Mar 19 '16 at 19:32

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