This is a detailed question about $U(N)$ intertwiners in LQG, and it comes from the the paper by Freidel and Livine (2011 - archive). It is very specific but related to finding a measure on a quotient space so I hope it will get more general mathematical physics interest.

We start by casting LQG into the spinor language. Most of the details are not necessary, but we have a set of $N$ spinors

$$|z_i\rangle =\left(\begin{array}{c} z^0_i\\ z^1_i\end{array}\right).$$

In addition to the usual spinor relations, they satisfy the closure relation, which is

$$\tag{23} \sum_i |z_i\rangle\langle z_i| =A(z)\mathbb{1}, \qquad A(z):=\frac{1}{2}\sum_i\langle z_i|z_i\rangle.$$

Now the states in the coherent intertwiner framework are holomorphic functions of $\mathbb{C}^{2N}$ that are homogenous functions under $GL(2,\mathbb{C})$ transformations. Thus, they are elements of a Grassmanian

$$\tag{81} Gr_{2,N}=\mathbb{C}^{2N}/GL(2,\mathbb{C}).$$

Now, we want a measure on this Hilbert space (and a resolution of unity) so we start by relating the measures on the three spaces above (Eq (93) in that paper):

$$\int_{\mathbb{C}^{2N}}\left(\prod_jd^4z_j \right)e^{-\sum_k\langle z_k|z_k\rangle}F(z)=\int_{GL(2,\mathbb{C})}\!\!\!\!\!\! d\lambda~ e^{-Tr(\lambda^\dagger\lambda)}|\det(\lambda)|^{2(N-2)}\int_{Gr_{2,N}}\!\!\Omega(z)F(z) .$$

1) Here is my first question; Why does the measure on $\mathbb{C}^{2N}$ have the exponential? Is it there because we are looking at coherent states? I am also a bit confused about the measure on $GL(2,\mathbb{C})$, but for the moment I'm ok just thinking it probably comes from the Haar measure and not the main point of my question.

The next thing they do in that paper is to say the measure on $Gr_{2,N}$can be represented as

$$\tag{94} \Omega(z)=\left(\prod_j d^4z_j\right) \delta^{(4)}\left(\sum_i|z_i\rangle \langle z_i|\right).$$

Then they go on to find the volume of $Gr_{2,N}$ as $\frac{\pi^{2(N-2)}}{(N-1)!(N-2)!}$ which gives a normalized measure on the Grassmanian.

2) So the next question is How does this measure jive with the closure constraints? If we perform the integration over the Grassmanian we enforce the constraint

$$\sum_{i}|z_i\rangle\langle z_i|=0\rightarrow A(z)=0.$$

While that doesn't directly contradict the closure constraints, it does seem like it is adding an extra condition which is not implied by anything else. Can someone help me understand this measure? (and extra points for anything about that $GL(2,\mathbb{C})$ part.)

  • $\begingroup$ The measure on $GL(2,\mathbb C)$ is the Lebesgue coming form $\mathbb C^4$. I wouldn't worry about the exponetial in the first integral. It's probably there because later they will compute integrals of that form. I am not sure I understand your last question. $\endgroup$ – MBN May 18 '13 at 8:40
  • $\begingroup$ The measure $\Omega$ has a delta function in it which sets $\sum |z_i\rangle \langle z_i|=0$ when the integral is performed. Why? It just seems to be adding something unnecessary. $\endgroup$ – levitopher May 18 '13 at 23:48
  • $\begingroup$ It is not unnecessary, without it the integral would be over $\mathbb C^N$ not the Grassmanian. $\endgroup$ – MBN May 19 '13 at 19:52

I) First note that there is a group action $\rho: GL(2,\mathbb{C})\times u(2) \to u(2)$ given by

$$\tag{A} g\quad \mapsto\quad\rho(g)\sigma~:= ~g\sigma g^{\dagger}, \qquad g\in GL(2,\mathbb{C}),\qquad\sigma\in u(2). $$

In detail, the Lie group

$$\tag{B} GL(2,\mathbb{C})~:=~ \{ g\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid \det(g)\neq 0\}$$

acts transitively on the set of Hermitian $2\times 2$ matrices

$$\tag{C} u(2)~:=~ \{\sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid \sigma^{\dagger}=\sigma \} ~=~ {\rm span}_{\mathbb{R}} \{\sigma_{\mu} \mid \mu=0,1,2,3\} .\qquad $$

Here $\sigma_{i}$, $i=1,2,3$ are the Pauli sigma matrices, and

$$\tag{D} \sigma_{0}~:=~{\bf 1}_{2 \times 2}$$

is the $2\times 2$ identity matrix.

Transitivity implies that the equation

$$\tag{E} \rho(g)\sigma_{0}~=~\sigma $$

has a solution $g\in GL(2,\mathbb{C})$ for arbitrary $\sigma\in u(2)$. Two solutions $g^{\prime},g^{\prime\prime} \in GL(2,\mathbb{C})$ to eq. (E) differ at most by a unitary transformation

$$\tag{F} (g^{\prime})^{-1}g^{\prime\prime} ~\in~ U(2). $$

Note in particular that $\sum_{i=1}^N|z_i\rangle \langle z_i| \in u(2)$ is generally a Hermitian $2\times 2$ matrix [even without applying the closure constraint (23)].

II) Let

$$\tag{G} d\mu(g)~=~\frac{d^8g}{|\det(g)|^4}, \qquad d^8g ~:=~ \bigwedge_{a,b\in\{1,2\}}d^2g^{a}{}_{b} ,$$ $$ d^2g^{a}{}_{b} ~:=~d({\rm Re}(g^{a}{}_{b})) \wedge d({\rm Im}(g^{a}{}_{b})), \qquad a,b~\in~\{1,2\}, \qquad g~\in~GL(2,\mathbb{C}), $$

be the two-sided positive Haar measure for the locally compact Lie group $GL(2,\mathbb{C})$.

III) Next consider the 4-dimensional real delta function

$$\tag{H} \delta^{4}\left(\rho(g^{-1})\sigma-\sigma_{0}\right).$$

on $u(2)$. It follows that the integral

$$\tag{I} {\cal N}~:=~\int_{GL(2,\mathbb{C})}\!\!\! d\mu(g)~\delta^{4}\left(\rho(g^{-1})\sigma-\sigma_{0}\right) ~<~\infty $$

is finite, cf. eq. (F), and that ${\cal N}$ does not depend on $\sigma \in u(2)$.

IV) Let $F(z)=F(gz)$ be an invariant function under the $GL(2,\mathbb{C})$ action

$$\tag{J} \mathbb{C}^2~\ni~ |z_i\rangle ~\to~ g|z_i\rangle~\in~\mathbb{C}^2, \qquad i\in \{1, \ldots, N\},$$

for $g\in GL(2,\mathbb{C})$ and $z=(z_1, \ldots, z_N)\in\mathbb{C}^{2N}$.

OP's first question concerns the choice of exponential weight factor. Here we will generalize the weight factor by introducing a real function $V:\mathbb{R}\to \mathbb{R}$. Ref. 1 chooses $V(x)=x$, but that is not necessary. Restrictions to the potential $V$ comes only from the non-compact nature of the integration, i.e. to keep the integrals finite.

Now calculate

$$\int_{\mathbb{C}^{2N}}\left(\prod_{j=1}^N d^4z_j \right)\exp\left[-{\rm tr}V\left(\sum_{k=1}^N|z_k\rangle \langle z_k|\right)\right] F(z)$$ $$~\stackrel{(I)}{=}~\int_{\mathbb{C}^{2N}}\left(\prod_{j=1}^N d^4z_j \right)\exp\left[-{\rm tr}V\left(\sum_{k=1}^N|z_k\rangle \langle z_k|\right)\right] F(z) $$ $$\times \int_{GL(2,\mathbb{C})}\!\!\! \frac{d\mu(g)}{{\cal N}}~\delta^{4}\left(\rho(g^{-1})\sum_{i=1}^N|z_i\rangle \langle z_i|-\sigma_{0}\right)$$ $$~\stackrel{|z_i\rangle=g|w_i\rangle}{=}~\int_{\mathbb{C}^{2N}}\left(\prod_{j=1}^N d^4w_j \right) \int_{GL(2,\mathbb{C})}\!\!\! \frac{d^8 g}{{\cal N}}~|\det(g)|^{2(N-2)} $$ $$\times \exp\left[-{\rm tr}V\left(\rho(g)\sum_{k=1}^N|w_k\rangle \langle w_k|\right)\right] \delta^{4}\left(\sum_{i=1}^N|w_i\rangle \langle w_i|-\sigma_{0}\right)F(w)$$ $$~=~\int_{GL(2,\mathbb{C})}\!\!\! \frac{d^8 g}{{\cal N}}~|\det(g)|^{2(N-2)} ~\exp\left[-{\rm tr}V\left(gg^{\dagger} \right)\right] $$ $$\times ~\int_{\mathbb{C}^{2N}}\left(\prod_{j=1}^N d^4w_j \right) \delta^{4}\left(\sum_{i=1}^N|w_i\rangle \langle w_i|-\sigma_{0}\right)F(w)$$ $$\tag{93'}~\stackrel{(94')}{=} ~\int_{GL(2,\mathbb{C})}\!\!\! \frac{d^8 g}{{\cal N}}~|\det(g)|^{2(N-2)} ~\exp\left[-{\rm tr}V\left(gg^{\dagger} \right)\right] ~\int_{Gr_{2,N}} \Omega(w)F(w),$$


$$\tag{94'} \Omega(w)~:=~\left(\prod_{j=1}^N d^4w_j\right) \delta^{4}\left(\sum_{i=1}^N|w_i\rangle \langle w_i|-\sigma_{0}\right).$$

So the answer to OP's second question is either:

  1. that Ref. 1 has a typo in eq. (94), where Ref. 1 has forgotten to subtract the unit matrix $\sigma_{0}$ in eq. (94), or

  2. that the subtraction of the unit matrix $\sigma_{0}$ in the delta function (H) is implicitly implied in the notation of Ref. 1.


  1. L. Freidel and E.R. Livine, $U(N)$ Coherent States for Loop Quantum Gravity, J. Math. Phys. 52 (2011) 052502, arXiv:1005.2090.

There is no necessity to start from the Gaussian measure on $\mathbb{C}^{2N}$. Any $U(N)$ invariant measure would result the same $U(N)$ invariant measure of the Grassmannian. However, this is the standard choice for two reasons.

1) When GL(2) is factorized out, the resulting measure on GL(2) is the "Ginibre measure" , Please see for example: Arxiv:1107.5019 (equation 3) by: Fischmann, Bruzda, Khoruzhenko, Sommers, and Zyczkowski. This is the measure of random matrices with complex gaussian entries. Also under the QR decomposition: $ g = qr$ $g \in GL(2), q \in U(2),r-\mathrm{Triangular}$, the induced measure on $U(2)$ is the Haar measure.

The second reason for this choice is that this measure on $\mathbb{C}^{2N}$ describes (the coherent states of) $2N$ harmonic oscillators. The procedure of constructing the Grassmannian according to this method is called "Kähler reduction". It is actually explained in equation 82 in the given article. This method allows the construction of interesting phase spaces from simple original ones; It is equivalent to Dirac's first class constraints.

Since in Qmechanic's answer, the error in the constraint equation is explained, I'll proceed to show the computation of the volume of the Grassmannian based on the constraint equations. I'll adopt the notation in which $Z$ is a $2\times N$ complex matrix composed of two complex $N$ vectors $u_1$ and $u_2$. Using this notation, the constraint equation components are given by:

$u_1^{\dagger}u_1 = 1$

$u_2^{\dagger}u_1 = u_1^{\dagger}u_2=0$

$u_2^{\dagger}u_2 = 1$

If the vector $u_1$ and $u_2$ were not constrained by the second equation, the first and the third equation describe the surface of $2N-1$ dimensional spheres. The second equation consists of the intersection of two hyperplanes at the origin, thus the solution of the constraint reduces one of the spheres to the $2N-3$ equator, thus the volume of the Grassmannian is given by:

$\mathrm{Vol}(Gr(2,N)) = \frac{1}{\mathrm{Vol}(U(2))}S_{2N-1}S_{2N-3}$

Where $S_k$ is the surface area of the $k$-sphere.

The division by the $U(2)$ volume is because the constraint is $U(2)$ invariant and all the $GL(2)$ dependence is taken care of in the first integral.

Taking into account:

$\mathrm{Vol}(U(2)) = \mathrm{Vol}(SU(2)) \mathrm{Vol}(U(1)) = S_{3}S_{1}$

The required volume is obtained from the substitution of the surface of odd dimensional spheres:

$S_{2k+1} = 2 \pi \frac{\pi^k}{k!}$

  • $\begingroup$ Great comment, I appreciate the details! Can I think of the division by the $U(2)$ volume as the Jacobian? As in normally one would have $du\to J(u)du$ for Jacobian $J(u)$ but that overcounts in this case because $u$ should be invariant under $U(2)$? $\endgroup$ – levitopher May 20 '13 at 20:41
  • 1
    $\begingroup$ @levitopher Yes, the $U(2)$ volume in this case is a component of the integration over $GL(2)$. In fact $Z$, or ( $u_1$ and $u_2$) are redundant for the Grassmannian, even with the constraints because they sum up to $2N-4$ coordinates, while the dimension of the Grassmannian is $2N-8$. Another way to look at it is that the constraints are gauge fixing conditions and $U(2)$ represents the gauge symmetry. This is exactly the meaning of equation 82 in the article. $\endgroup$ – David Bar Moshe May 21 '13 at 7:12
  • 1
    $\begingroup$ @levitopher cont. In fact, the procedure adopted by Qmechanic is a part of the "Faddeev-Popov" procedure for reducing gauge symmetries, and in principle it can be completed to get a BRST invariant integrand. $\endgroup$ – David Bar Moshe May 21 '13 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.