# How are the intertwiners related to Schur's lemma?

• A
• Heidi
In summary, the author of the paper describes a decomposition of the tensor product of two representations of the rotation group, and shows that total angular momentum must sum to zero at a spin foam vertex.

#### Heidi

Hi Pfs
there are several Schur's lemma.
i am talking about the one saying this one:

Let E be a finite dimensional vector space over C and U an irreducible subset of L(E). If an endomorphism ϕ of E commutes with any element of U, then ϕ is a dilation

I consider a vertex in a spinfoam with two incoming edges and one outgoing one.
we have SU2 irreps on D E F associated to these legs
if i have two matrices acting on D and E, an intertwiner will map them to a third one acting on F.
how to relate this to the Schur's lemma with only one vector space and a dilation.
i think that the vector space contains the tensor product of E and F and maybe F but i do not see how to get the map on this space.
thanks

Having skimmed the figures in several (all right, two) spin foam papers, isn't an intertwining operator essentially an angular momentum recouping or some Clebsch Gordan matrix thing. Like a Wigner-3j? Maybe look at how the Wigner-Eckart theorem is constructed?

i found a very explicit paper:
https://www.imperial.ac.uk/media/im...tations/2014/Clement-Delcamp-Dissertation.pdf
but there a details that i do not understand.
look at page 23
we have two incoming edges labelled by 3/2 and an outgoing one labelled by 2.
the author make a decomposition of the tensor product of the 3/2 representations and look for an invariant subspace
he says that it is V^2 (why this one and not V^1)
and that the dimension of this subspace is equal to one. how to see that?

Okay, I think I get it. These spin foam diagrams are like Feynman diagrams, they represent a matrix. The external edges are labeled by the quantum numbers of the incoming states and the outgoing states. Space-time is homogeneous and isotropic in the usual Feynman case, so one of the labels is the 4-momentum of the particle. No matter what goes on with the non-momentum quantum numbers at the vertex, total 4-momentum is conserved, or, the total 4-momentum sums to 0. If we have 2 incoming lines and one out, ##p_1+p_2 = p_3##, or

##p_1+p_2-p_3=0##.

At a spin foam vertex graph total angular momentum is conserved no matter what goes on with the color degrees of freedom. Here is where Wigner-Eckart aka Schur's lemma, comes in. In the diagram, the total angular momentum must sum to zero. The color operators must respect this sum and commute (be dilations in your terminology) with all the rotation parts of this matrix.

Now, the ##V^s## are ##2s+1## dimensional spaces because the ##s## is the associated spin of the space. Clearly, ##V^0## is a one dimensional space. If we look at the ##3/2\otimes 3/2\rightarrow 2## part of the matrix, we need the,

## \mathbf{\frac{3}{2}} \otimes \mathbf{\frac{3}{2}}\otimes \mathbf{2} \rightarrow \mathbf{0}##

part. This is the one dimensional ##V^0## bit. The point is, on this subspace the color(s) operator must be constant in order to maintain rotational invariance. And, the same must hold for the ##\mathbf{0}, \mathbf{1}##, and ##\mathbf{3}## outgoing spin cases.

Heidi
many things are srill obscure to me:
if the two incoming spins are 3/2 how can there be conservation on a 2 representation like on the picture?
and what is the meaning of 3/2 \otimes 3/2 \otimes 2 ---> 0
same question for 0 1 and 3?
how to text latex formulas before posting the reply

See the "Latex Guide" on the leftmost side at the bottom of the last post in any thread. I believe that PF on phones works alright with Latex.

Heidi
Heidi said:
many things are srill obscure to me:
if the two incoming spins are 3/2 how can there be conservation on a 2 representation like on the picture?
and what is the meaning of 3/2 \otimes 3/2 \otimes 2 ---> 0
same question for 0 1 and 3?
how to text latex formulas before posting the reply
Well, that's a lot of ground to cover. None of it is very hard, actually, just a lot of typing. Perhaps you should look at a text on angular momentum? These cover the representations of the rotation group from the perspective of physical applications. My recommendation is "Angular Momentum" by D. M. Brink and G. R. Satchler, second edition. It's a small book. It also has a chapter on graphical methods which I think is particularly relevant here. Without these basics in hand, the thesis you're reading will make little sense.

are there other books which are in the kindle format?
or on the web?
i would appreciate a self contained course with the graphical interpretation of intertwiners.
i am visually impaired , i can not read articles on paper, only on screen. (i invert the colors)
thanks
Edit: it is obvious that my aim is not to study a thesis about entanglement in networks. i know my level. i gave a link to that text because i did not find simpler ones with the graphical interpretation of intertwiners. understanding this point would be enough

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Sorry, I don't know of such a text. Also, my analogy with Feynman diagrams may be a bit off kilter. In one of the reviews I thumbed through a remark was made that the similarity with Feynman diagrams is "striking" but it seems that's not where spin foams are coming from. Also, Equation 1.53 (I think) reads ##j_1+j_2+j_3=N## for some integer ##N##. So summing to zero at each vertex may also be off. On the other hand, angular momentum coupling is everywhere to be found in the paper so learning something about it will be useful. Best of luck in your studies. Try searching arXiv.org for spin foam reviews. I got a bunch of hits.

Heidi and Paul Colby
my question came from a remark. i think it was like that:
giving the trivalent intertwiner from V1 tensor V2 to V3 is equivalent to
giving an homothecy on V1 tensor V2 tensor V3*
i do not exactly remember what was the third term (maybe the forms upon V3?)
so the idea was to use the Schur's lemma on this new space.
i wondered too if the homothécy can be "seen" in the initial intertwiner.

Could you tell me if this idea is good:
when i have an incoming edge E1 and an outgoing edge E2 in the trivalent case we have an intertwiner between "In" Hilbert spaces and "out" space
the graphical interpretation shows lines coming from E1 which go to E2
if we take one of these oriented lines it will follow a loop which will retrun to E1.
we had an holonomy and we get a wilson loop
So when we had a homomorphism from V1 to V2 the loop give an invariant map from V1 to itself.
may it be where Schur's lemma can be applied?

I don't know. I'm still working my way through the primer Demystifier provided. My plan is to stay the course and understand what is before I ask about what might.

Heidi
my question is exactly: how to derive eq 68 in the primer given by demystfier.

I actually think (68) is a typo. ##\hom_G(V,W)## are maps from ##V## to ##W##. (68) should read,

##\hom_G(V,W) \simeq \text{Inv}_G(W\otimes V^\ast)##​

if I understand correctly. If we write out an element of ##\hom_G(V,W)## in Dirac notation they look like,

##T = \sum_{v\in V}\sum_{w\in W} T_{w,v} |w\rangle\langle v|##,​

where, ##T_{w,v}## is a matrix of complex numbers, ##|w\rangle\in W## and ##\langle v|\in V^\ast##. Clearly, ##|w\rangle\langle v|\in W\otimes V^\ast##. Now, and this is the important bit, we want to select those matrices that are ##G## invariant. If ##G## is the rotation group, then we want to select those one dimensional subspaces that are constant for all rotations. If we split the two spaces, ##W## and ##V##, up into invariant subspaces, then one is looking for all the ##G## or rotationally invariant bits.

For the trivalent case, the in-space are all possible couplings of spin ##3/2## with ##3/2##. This coupling gives ##0##, ##1##, ##2## and ##3##.

This is as far as I've gotten. How the diagrams with multiple internal lines are intended is still a mystery to me. They seem to be including a line for every half integral spin in the incoming and outgoing legs. Beats me.

I saw this written differently
it was said that having an intertwiner from ## V1 \otimes V2## to W
was equivalent (giving the same information) that having
## V1 \otimes V2 \otimes W^* ## mapped to a complex number in C
it is the homothecy given by Schur's lemma and we have global invariance on given space not unchanged vectors like in the link's definition.

Heidi said:
I saw this written differently
Yes, if the vector spaces are finite dimension one may freely exchange ##V^\ast## for ##V## since they are isomorphic. To me the logic is backward. Mapping from ##V## takes a ##V^\ast## when one writes out the operators as matrices. Writing it out the way you've seen it takes an extra isomorphism. Doesn't hurt anything but it clutters the development up with unneeded steps, IMO. On the other hand, maybe this freedom adds degrees of freedom one needs for other reasons. I just don't know.

Heidi said:
it was said that having an intertwiner from V1⊗V2 to W
was equivalent (giving the same information) that having
V1⊗V2⊗W∗ mapped to a complex number in C
Yes, very same could be said for ##V1^\ast\otimes V2^\ast\otimes W## since all vector spaces are finite dimensional in the things I've been reading. The advantage of writing my way is I don't have to reverse all my incoming and outgoings. I don't believe there is a substantive difference.

Heidi said:
it is the homothecy given by Schur's lemma and we have global invariance on given space not unchanged vectors like in the link's definition.
These intertwining operators are just matrices. Matrices are themselves a vector space that is acted upon by ##G##. We may reduce this space the same way we reduced the row and column spaces. The ##G##, or rotationally invariant bit, are those reduced matrix bits that are constant under action of ##G## or rotations. There are in general many ways to couple (reduce) this matrix space. Each way involves another constant the user is free to choose.

there is alson Qunatum Gravity by Rovelli
http://www.cpt.univ-mrs.fr/~rovelli/book.pdf
see the appendix at page 277
the case of the theta spin network is treated there.
there is an horizontal spin = 1 edge oriented to the right
and two half circle edges oriented to the left.
i wonder if a change in the orientation of the horizontal edge gives another spin nerwork.
what are the formulas for this new network with to nodes , one with three outgoing edge and with three ingoing ones. what is the total area around them, is it null?

So I fired up my computer algebra program to do some checking. Equations A.15 and A.18 aren’t consistent without some additional help. The Wigner 3js are real numbers while equation A.15 clearly involves complex numbers. Okay, this is where things get sloppy. One has to know the ##i## index in A.15 refers to Cartesian coordinates while the ##i## in A.18 refers to spherical tensor components. This isn’t a big deal but it’s the kind of thing that can derail one’s understanding.

Hi Paul,
something could me help to understand the problem:
considere an intertwinet mapping ##C \otimes D## to ##E \otimes E##
we have here a node with two incoming edges and two outgoing ones.
now i reverse the direction of one outgoing edge to become ingoing.
How to construct the new intertwiner mapping three hilbert spaces to F?

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Okay, sure. ##C## is a label. Let's make it work for us. Let ##C## be some spin, ##\{0,1/2,1,3/2,\cdots\}##. This means ##C## is spanned by ##2C+1## vectors. The space, ##C\otimes D## has ##(2C+1)(2D+1)## vectors. The space ##E\otimes E## has ##(2E+1)^2## vectors. For most choices of ##E## the intertwine operator will be identically 0. For the cases where it's not zero, there may be many possible ways to map to this space. Let's count the possible ways.

First off ##C\otimes D## may be coupled to ##W## where ##|C-D| \le W \le |C+D|## where we only allow ##W## values that differ by an integer. Now, ##W## must also be ##0 \le W \le 2E##. Okay, so ##W## must be an integer. This means if ##C=1/2## and ##D=1##, the intertwine operator is 0.

Let's choose ##C=1##, ##D=2## and ##E=1/2##. In this case only ##W=1## is possible. The intertwine operator matrix will be just the Wigner-3j.

Moving on, if there are three incoming spins, ##C, D##, ##E## there are more options. We may first couple ##C## and ##D## to ##W## and then couple ##W## to ##E## to get ##F##. Or we may couple ##D## and ##E## to ##W## then couple ##W## to ##C## to get ##F## ... and so on.

Each of these couplings introduces a Wigner-3j. Each provides an intertwining operator and is associated with some constant that one is free to chose.

Heidi
thanks Paul
i wanted to map C,D to E,F (not to E,E)
and then to map something to F
is it harder?
my aim is to understand what happens then when all the edges of a node are incoming.

i think that there is a more "mechanical" way to get the second intertwiner
from the first one which maps C,D to E,F
there is a graphical interpretaition to such a 4 valent intertwiner
the number on lines on the edges does not depand of the orienation of the edges. the orientation of an outgoing edge simply gives the incoming lines to the "next" node
if we reverse the orientation of an outgoing edge it decreases the number of "next" nodes. here we had E,F and after we have only F
what about the graphical interpretation of the new case?