User aram - MathOverflow

injective tensor norms for real tensors

2011-09-30T22:59:24Z

If $A$ is an element of $\mathbb{R}^n \otimes\mathbb{R}^n \otimes\mathbb{R}^n$, then define its injective tensor norm to be $$\|A\|_{\rm inj} := \max_{x,y,z\in \mathbb{R}^n, \|x\|=\|y\|=\|z\|=1} |\langle A, x\otimes y\otimes z\rangle|.$$ Here the norm on vectors is the usual Euclidean norm.

I have two questions.

Does this norm change if x,y,z can be taken to arbitrary complex unit vectors?
What if $A$ is symmetric under exchange of the first two positions? That is,
$\langle A, x\otimes y\otimes z\rangle = \langle A, y\otimes x\otimes z\rangle$ for all $x,y,z$.

Answer by aram for injective tensor norms for real tensors

2011-10-01T01:02:47Z

I just found this paper, which gives an example in which the real and complex version of the norm are different. The tensor in this example is also symmetric, which provides an example for part 2 as well.

Answer by aram for Quantum analogue of Wiener process

2011-06-27T16:23:23Z

In section III.B of the survey paper you cite, it describes continuous quantum walks, which are I think are a natural analogue of the Wiener process. These are basically Hamiltonian evolution when the Hamiltonian is something like the adjacency matrix (or Laplacian) of a graph.

On the relationship between continuous- and discrete-time quantum walk has some recent developments with fascinating applications to simulating Hamiltonians on quantum computers.

Answer by aram for Does BQP^P = BQP ? ... and what proof machinery is available?

2011-01-27T17:33:10Z

Yes, P is contained in BQP (Benioff, 1982; http://prl.aps.org/abstract/PRL/v48/i23/p1581_1 ) and $BQP^{BQP}=BQP$, for pretty much the same reason $BPP^{BPP}=BPP$. This second point first appeared (that I know of) as Cor 4.15 of BBBV'97: http://www.cs.berkeley.edu/~vazirani/pubs/bbbv.ps .

Comment by aram

2011-10-01T00:58:48Z

Yes. Sorry for being unclear.

Comment by aram

2011-09-30T23:01:52Z

I'm having trouble getting the $$ equation to work.