User aram - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:41:21Z http://mathoverflow.net/feeds/user/7718 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76894/injective-tensor-norms-for-real-tensors injective tensor norms for real tensors aram 2011-09-30T22:59:24Z 2011-10-02T17:43:53Z <p>If $A$ is an element of $\mathbb{R}^n \otimes\mathbb{R}^n \otimes\mathbb{R}^n$, then define its <a href="http://en.wikipedia.org/wiki/Topological_tensor_product" rel="nofollow">injective tensor norm</a> to be <code>$$\|A\|_{\rm inj} := \max_{x,y,z\in \mathbb{R}^n, \|x\|=\|y\|=\|z\|=1} |\langle A, x\otimes y\otimes z\rangle|.$$</code> Here the norm on vectors is the usual Euclidean norm.</p> <p>I have two questions.</p> <ol> <li>Does this norm change if x,y,z can be taken to arbitrary <i>complex</i> unit vectors?</li> <li>What if $A$ is symmetric under exchange of the first two positions? That is,<br> $\langle A, x\otimes y\otimes z\rangle = \langle A, y\otimes x\otimes z\rangle$ for all $x,y,z$.</li> </ol> http://mathoverflow.net/questions/76894/injective-tensor-norms-for-real-tensors/76902#76902 Answer by aram for injective tensor norms for real tensors aram 2011-10-01T01:02:47Z 2011-10-01T05:47:50Z <p>I just found <a href="http://dmle.cindoc.csic.es/pdf/RRACEFN_2000_94_04_05.pdf" rel="nofollow">this paper</a>, which gives an example in which the real and complex version of the norm <i>are</i> different. The tensor in this example is also symmetric, which provides an example for part 2 as well.</p> http://mathoverflow.net/questions/15973/quantum-analogue-of-wiener-process/68944#68944 Answer by aram for Quantum analogue of Wiener process aram 2011-06-27T16:23:23Z 2011-06-27T16:23:23Z <p>In section III.B of the survey paper you cite, it describes <em>continuous quantum walks</em>, 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.</p> <p><a href="http://arxiv.org/abs/0810.0312" rel="nofollow">On the relationship between continuous- and discrete-time quantum walk</a> has some recent developments with fascinating applications to simulating Hamiltonians on quantum computers.</p> http://mathoverflow.net/questions/53503/does-bqpp-bqp-and-what-proof-machinery-is-available/53513#53513 Answer by aram for Does BQP^P = BQP ? ... and what proof machinery is available? aram 2011-01-27T17:33:10Z 2011-01-27T17:33:10Z <p>Yes, P is contained in BQP (Benioff, 1982; <a href="http://prl.aps.org/abstract/PRL/v48/i23/p1581_1" rel="nofollow">http://prl.aps.org/abstract/PRL/v48/i23/p1581_1</a> ) 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: <a href="http://www.cs.berkeley.edu/~vazirani/pubs/bbbv.ps" rel="nofollow">http://www.cs.berkeley.edu/~vazirani/pubs/bbbv.ps</a> .</p> http://mathoverflow.net/questions/76894/injective-tensor-norms-for-real-tensors Comment by aram aram 2011-10-01T00:58:48Z 2011-10-01T00:58:48Z Yes. Sorry for being unclear. http://mathoverflow.net/questions/76894/injective-tensor-norms-for-real-tensors Comment by aram aram 2011-09-30T23:01:52Z 2011-09-30T23:01:52Z I'm having trouble getting the  equation to work.