The n-Category Café

I will write next time in which sense $a_1+a_1^*$ and $a_2+a_2^*$ are free; let me for now just comment on whether one can impose commutation relations between the annihilation operators.

As I said before, the q-commutation relations together with the fact that the vacuum is annihilated by the annihilation operators determines all moments of the $a_i$ and $a_j^*$ with respect to the vacuum expectation state, and for general $q$ these are not compatible with any kind of relation between $a_i$ and $a_j$. For $q=+1$ it turns out that the vacuum expectation vanishes on the ideal generated by $a_i a_j - a_j a_i$, thus we can add the relation $a_i a_j -a_j a_i =0$ without any harm; for $q=-1$ it turns out that the vacuum expectation vanishes on the ideal generated by $a_i a_j + a_j a_i$, thus we can add the relation $a_i a_j+ a_j a_i=0$. For all the other $q$, however, there is no such ideal which vanishes under the vacuum expectation, so in all those cases (including the “free” case $q=0$) there are no relations between the annihilation operators.

[The same can be rephrased in the following way: whereas the relations for $q=1$ can be realized on the symmetric Fock space and for $q=-1$ on the anti-symmetric Fock space, for all the other $q$ we have realizations on the full Fock space without dividing out anything (in particular, without using any form of q-deformed symmetrization).]

So in particular, for $q=0$ this means that our free operators satisfy no algebraic relation and are free in the algebraic sense. However, freeness in the sense of free probability is a much stronger condition than just that; in a similar way as independence of two random variables does not just mean that they commute, but is a very precise condition on moments of the variables. I will say something more precise on freeness next time.

Roland wrote:

whereas the relations for $q=1$ can be realized on the symmetric Fock space and for $q=−1$ on the anti-symmetric Fock space, for all the other $q$ we have realizations on the full Fock space without dividing out anything (in particular, without using any form of $q$-deformed symmetrization).

Okay, great — that’s very clear, especially since I know that the ‘full Fock space’ is what mathematicians call ‘the tensor algebra’ $T V$ on a vector space $V$ (or the Hilbert space completion of $T V$, if $V$ is a Hilbert space).

But, I’m also fond of certain $q$-deformed Fock spaces that we get when $V$ is equipped with a Hecke $R$-matrix, that is, an operator

$$R: V \otimes V \to V \otimes V$</p> <p>satisfying the <b>Yang--Baxter equations</b></p> <p>$$(R \otimes 1)(1 \otimes R)(R \otimes 1) = (1 \otimes R)(R \otimes 1)(1 \otimes R)$$</p> <p>and the <b>Hecke relations</b> --- some quadratic equation ensuring that$R^2$has just two eigenvalues, say$q$and$-q^{-1}$, though nobody can agree on the precise normalization. These relations let us define `$R$-symmetrization' and then an `$R$-symmetrized Fock space'. <a href = "http://www.google.com/search?q=q-deformed+fock+space">A lot of people</a> think about things like this.</p> <p>Can your$q$-deformed annihilation and creation operators be represented on some$q$-deformed Fock space, or only on the full Fock space? </p> <p>(I'm just trying to see if free probability theory is connected to this$q$-deformed Fock space game, or not.)