Let be a large natural number, and let
be a matrix drawn from the Gaussian Unitary Ensemble (GUE), by which we mean that
is a Hermitian matrix whose upper triangular entries are iid complex gaussians with mean zero and variance one, and whose diagonal entries are iid real gaussians with mean zero and variance one (and independent of the upper triangular entries). The eigenvalues
are then real and almost surely distinct, and can be viewed as a random point process
on the real line. One can then form the
-point correlation functions
for every
, which can be defined by duality by requiring
for any test function . For GUE, which is a continuous matrix ensemble, one can also define
for distinct
as the unique quantity such that the probability that there is an eigenvalue in each of the intervals
is
in the limit
.
As is well known, the GUE process is a determinantal point process, which means that -point correlation functions can be explicitly computed as
for some kernel ; explicitly, one has
where are the (normalised) Hermite polynomials; see this previous blog post for details.
Using the asymptotics of Hermite polynomials (which then give asymptotics for the kernel ), one can take a limit of a (suitably rescaled) sequence of GUE processes to obtain the Dyson sine process, which is a determinantal point process
on the real line with correlation functions
where is the Dyson sine kernel
A bit more precisely, for any fixed bulk energy , the renormalised point processes
converge in distribution in the vague topology to
as
, where
is the semi-circular law density.
On the other hand, an important feature of the GUE process is its stationarity (modulo rescaling) under Dyson Brownian motion
which describes the stochastic evolution of eigenvalues of a Hermitian matrix under independent Brownian motion of its entries, and is discussed in this previous blog post. To cut a long story short, this stationarity tells us that the self-similar -point correlation function
obeys the Dyson heat equation
(see Exercise 11 of the previously mentioned blog post). Note that vanishes to second order whenever two of the
coincide, so there is no singularity on the right-hand side. Setting
and using self-similarity, we can rewrite this equation in time-independent form as
One can then integrate out all but of these variables (after carefully justifying convergence) to obtain a system of equations for the
-point correlation functions
:
where the integral is interpreted in the principal value case. This system is an example of a BBGKY hierarchy.
If one carefully rescales and takes limits (say at the energy level , for simplicity), the left-hand side turns out to rescale to be a lower order term, and one ends up with a hierarchy for the Dyson sine process:
Informally, these equations show that the Dyson sine process is stationary with respect to the infinite Dyson Brownian motion
where are independent Brownian increments, and the sum is interpreted in a suitable principal value sense.
I recently set myself the exercise of deriving the identity (3) directly from the definition (1) of the Dyson sine process, without reference to GUE. This turns out to not be too difficult when done the right way (namely, by modifying the proof of Gaudin’s lemma), although it did take me an entire day of work before I realised this, and I could not find it in the literature (though I suspect that many people in the field have privately performed this exercise in the past). In any case, I am recording the computation here, largely because I really don’t want to have to do it again, but perhaps it will also be of interest to some readers.
The basic tool here is cofactor expansion, which we write as follows. Given an matrix
, we can expand the determinant
as
where is the bottom left row of
,
is the top left minor of
, and
is the
matrix formed by replacing the
row of
by
.
As an example of how this cofactor expansion is used in determinantal point processes, let us recall Gaudin’s lemma, which we state slightly informally:
Lemma 1 (Gaudin’s lemma) Let
be a (sufficiently nice) function obeying the idempotent relation
For each
, let
be the correlation function
Then we have
Proof: From (4) we have
where and
. Integrating using (6), we conclude that
By the multilinearity of determinant, we can write
But from (5) we have
and so
and the claim follows.
To establish (3), we will need some identities that are specific to the sine kernel (2) that are reminiscent of (5) and (6). The analogue of (6) is very easy:
The sine kernel actually does obey (5), but we will need the following more general identity:
Lemma 2 For any real numbers
, one has
(using the principal value interpretation of the integral) when
are distinct. In particular, by l’Hopital’s rule one has
Note that if one multiplies by in (9) and then sends
, one recovers (5).
Proof: This can be verified by a routine contour integration, but we will give a more Fourier analytic proof instead. We will omit some (routine) details concerning justifying the various interchanges of integrals we will be using here. By a limiting argument we may take to be distinct. Observe that
and so the left-hand side of (9) can be rewritten as
From the basic identity we have
for any , so the preceding expression simplifies to
We split this as the difference of
For (12), we make the change of variables and rewrite this as
We have (in the distributional sense at least) that
and so by (9) this term simplifies to .
Now we turn to (13). Here we observe that in the region of integration we have , and so this expression can be factored as
The final factor here is thanks to (9), while a brief computation shows (in the distributional sense, at least) that
and the claim follows.
Now we turn to the verification of (3). It will suffice to establish the identity
for each . By a continuity argument we may take the
to be distinct.
Fix . By (7), the left-hand side of (14) can be written as the difference of
and
The first expression vanishes thanks to (8), so we turn to the second. From Lemma 2 we have
when , and
when . In the former case, we see from row operations that
while from the Leibniz rule and symmetry (and (8)) we have
putting all of these facts together, we obtain (14) as required.
Remark 1 Presumably, a similar exercise can be performed for the Airy process that governs the asymptotics of extreme eigenvalues of GUE (which include the famous Tracy-Widom law), but I did not attempt this.
10 comments
Comments feed for this article
11 November, 2012 at 1:53 pm
ANDI
Interesting! Well done!
11 November, 2012 at 3:08 pm
Geoffrey Irving
Unfortunately the formatting for this post is very distorted for me: http://naml.us/random/tao-formatting.png. It might be due to a Mac Retina display issue. Do you know if anyone else has seen this?
11 November, 2012 at 3:09 pm
Geoffrey Irving
And…it just fixed itself. Odd.
11 November, 2012 at 3:25 pm
Anonymous
“it just fixed itself” –> “it just got fixed” ;)
12 November, 2012 at 6:45 am
mfrasca
After the sentence “…are the (normalised) Hermite polynomials; …” I think a link is missing. [Corrected, thanks – T.]
13 November, 2012 at 8:22 pm
Anonymous
I see properly formatted equations until two equations past eq. (11), where it says “From the basic identity…”, the dispay equation after that, and all later equations, are distorted.
24 November, 2012 at 10:52 pm
yypus
In (10), x-t should be y-t. Any one can explain or refer to a book why the two displays are true after “(in the distributional sense, at least)”?
25 November, 2012 at 8:30 am
Terence Tao
Thanks for the correction! These identities are essentially equivalent to the fact that the Hilbert transform is a Fourier multiplier with symbol
(depending on one’s sign conventions). Note that these integrals can also be evaluated in the principal value sense by contour integration, and with some standard truncation arguments one can show that the principal value sense of the integral matches up with the distributional interpretation of the integral.
26 November, 2012 at 10:07 am
Mike J
Terence, I have a completely out-of-office question. I have a truly marvelous reason for asking, which this box is too narrow to contain. I could think of no more appropriate place to ask than here, though even here it is still far from appropriate – I can only hope I catch you in a fine and humorous mood.
Do you drink coffee?
27 January, 2013 at 10:58 pm
Thomas Watson
Good question. I am curious too.