0
$\begingroup$

While reviewing my course in QFT, I find that I am having a bit of trouble understanding how we define right and left action representations (that's probably because they are not defined). Let me try and show what I have understood (it might be very well be wrong) :

I will use the $\cdot$ to denote the usual matrix multiplication (i.e. if $A$ and $B$ are matrices, then $A\cdot B$ is a matrix product). I know this is unusual, but I sense it will get messy if I don't have an explicit symbol.

Let $\rho : G\rightarrow Aut(V)$ be a finite dimensional representation of a group G on a vector space V. Since $\rho$ acts on $V$ with a linear action, we can represent this action by the left-action of a matrix, defined as $\rho(g)(v) = \rho^{L}(g)\cdot v$ where $\rho^{L}(g) \in GL(V)$. $\rho^{L}$ is what I understand to be the left-action representation. By imposing the representation property on $\rho$, we find that $\rho^{R}(g)\cdot\rho^{L}(h) = \rho^{L}(gh)$

However, the linear action of $\rho(g)$ on $V$ can just as well be represented as a right action of a matrix, defined as $\rho(g)(v) = v\cdot\rho^{R}(g)$. Again, $\rho^{R}(g)$ is what I understand to be the right-action representation. The representation property then gives us $\rho^{R}(g)\cdot\rho^{R}(h) = \rho^{R}(hg)$.

So, as I understand it, the left and right representation are just two ways of writing a representation of a group $G$. My first question is, is my understanding correct ?

EDIT : Here, I considered the matrix representation of $\rho^L(g)$ and $\rho^R(g)$, but we could be more general and define an abstract action using linearity of $GL(V)$ and action on a basis of V.

My second question heavily relies on this first part, so I will edit it if it turns out that my understanding is wrong.

How do we establish an "equivalence" between a right action, and a left action representations ? I know that for two representations $\rho$, $\rho'$ to be equivalent we need to have an bijective intertwiner between the representation. But what if we decide to implement $\rho$ as a right action, and $\rho'$ as a left action ? Can we still compare them using this intertwiner definition ?

Example : Consider a left-action representation $\rho(g)(v) = \rho^L(g)\cdot v$ and the right action representation induced by the previous one $\rho'(g)(v)=v^T\rho^L(g)^T$, where I wrote $v^T$ to emphasize that it is a row vector. In this case I think the representations are equivalent since if we choose the linear map $\phi(v) = v^T, \phi : V\rightarrow V^T$ we have that $$\phi(\rho(g)(v))=\phi(\rho^L(g)\cdot v) = v^T \rho^L(g)^T = \phi(v)\rho^L(g)^T=\rho'(g)(\phi(v))$$

Now, we can also consider another representation constructed from $\rho$ using the inverse : $\rho''(g)(v) = v^T \rho^L(g)^{-1}$. For this one, I haven't been able to construct a proper intertwiner that shows that it is equivalent to $\rho(g)$. However, I really think that they should be, because of the way they are constructed. This is one of the reasons why I am unsure whether I can use the definition of "equivalence" between left and right representations.

Since I am not sure whether the whole problem is clear, I'd like to motivate how these questions came up.

Motivation: In QFT, we have left and right spinors, $\psi_L$ and $\psi_R$, which transform respectively in the $(1/2,0)$ and $(0,1/2)$ representations of $SL(2,\mathbb{C})$, which we will call $\rho_-$ and $\rho_+$ respectively. Furthermore, we have that $\rho_- = (\rho_+^\dagger)^{-1}$. Both are implemented as left-action.

Now consider $g\in SL(2,\mathbb{C})$, then, under a lorentz transformation $\psi_L$ transforms in the $\rho_-$ representation of $SL(2,\mathbb{C})$ as $\psi_L\rightarrow \rho_-(g)\psi_L$. Taking the conjuguate transpose we find the transformation law of $\psi_L^\dagger \rightarrow \psi_L^\dagger \rho_-(g)^\dagger$. As we can see, $\psi_L^\dagger$ transforms in some representation implemented as a right-action. Now, we claim that $\psi_L^\dagger$ transforms in the $(0,1/2)$ representation, that is it transforms in an equivalent representation as $\rho_+$. Thus, if my understanding is correct, we should be able to show that the right-action representation acting on $\psi_L^\dagger$ is equivalent to the left-action representation acting on $\psi_R$, and hence all my questions.

$\endgroup$
  • $\begingroup$ If you wish to literally think of $\rho^L(g)$ as a matrix, then in order for $\rho^L(g) \cdot v$ to be defined you must literally think of $v$ as a column matrix. But if you wish to literally think of $\rho^R(g)$ as a matrix, then in order for $v \cdot \rho^R(g)$ to be defined you must literally think of $v$ as a row matrix. And you can't do both at the same time. What you could do, on the other hand, is to think of $v$ as a column matrix so $\rho^L(g) \cdot v$ makes sense, and then its transpose $v^T$ is a row matrix, so $v^T \cdot \rho^L(g)$ now makes sense. $\endgroup$ – Lee Mosher Dec 9 '17 at 18:13
  • $\begingroup$ And when I write "transpose", I suppose that I really mean "conjugate transpose" since you are working over the complex numbers. $\endgroup$ – Lee Mosher Dec 9 '17 at 18:14
  • $\begingroup$ Yes, that is what I meant to say, in both my definitions of the right and left reps I use the same $v$, but one is a column vector and the other is a row vector. I didn't want to use transpose explicitely since it could work also with the conjuguate transpose, for example. This leads me to another problem, I don't understand why we should consider the conjuguate transpose ? What prevents me from consider the transpose even though the rep is complex ? $\endgroup$ – Frotaur Dec 9 '17 at 18:18
1
$\begingroup$

I would like to stay away from transposing matrices/vectors, or multiplying vectors on the left of matrices, since I think that it obscures what is really going on, which is that a left representation preserves the order of multiplication, while a right representation reverses it.

You would have learnt that a left representation of $G$ on $V$ is a group homomorphism $\rho^L: G \to \mathrm{GL}(V)$, that is, a map of sets such that $\rho^L(gh) = \rho^L(g) \rho^L(h)$. If we write $g \cdot v$ as short notation for $\rho^L(g) v$, then a left representation satisfies $gh \cdot v = g \cdot (h \cdot v)$.

What then, is a right representation? It should be some map of sets $\rho^R: G \to \mathrm{GL}(V)$, and if we write $v \cdot g$ as notation for $\rho^R(g)v$, then we want it to satisfy $(v \cdot g) \cdot h = v \cdot gh$, which when written in terms of the $\rho^R$ is the condition $\rho^R(h) \rho^R(g) = \rho^R(gh)$. So a right representation is a group antihomomorphism $\rho^R: G \to \mathrm{GL}(V)$, which reverses the order of multiplication.

I'm not sure of what to say about comparing right and left representations, since the reversing of order messes up the usual definition of intertwiner. However, there are some ways to get a left representation from a right one.

Starting with a right representation $\rho^R: G \to \mathrm{GL}(V)$, we can precompose it with the map $\mathrm{inv}: G \to G$, $\mathrm{inv}(g) = g^{-1}$. Then $\rho^R \circ \mathrm{inv}$ will be a homomorphism, and so a representation. Note that $\mathrm{inv}$ is an anti-homomorphism, and the composition of two anti-homomorphisms will be a homomorphism.

Instead of reversing order on the group side using an inverse, we could reverse it on the vector space side, by using a transpose. So if $\mathrm{trans}: \mathrm{End}(V) \to \mathrm{End}(V)$ denotes taking matrix transpose (or conjugate transpose), then $\mathrm{trans} \circ \rho^R$ will be a left representation. (The order reversing happens via the transpose: $(AB)^T = B^T A^T$).

$\endgroup$
  • $\begingroup$ Thank you for the answer, it does clear up some of my confusing. Indeed, I agree that passing through vectors and matrices isn't ideal, but at the time I didn't see how to best define it. However, I still think there should be some way of compare and "connect" right and left representations. Indeed, people in the QFT literature do state that $\psi_R^\dagger$ transforms in the same rep as $\psi_L$, so there must be a way do compare them, or maybe it is just some physicist sloppy work, and what we mean is not an actual isomorphism of representations ? $\endgroup$ – Frotaur Dec 9 '17 at 23:34
  • $\begingroup$ Well, if $\psi_R$ is a right representation, then by what I said above about the transpose, $\psi_R^\dagger$ would be a left representation, so then you could compare them as left representations. $\endgroup$ – Joppy Dec 9 '17 at 23:49
  • $\begingroup$ Sorry maybe I wasn't very clear, but when I considered $\psi_R^{\dagger}$ I meant it as a row vector. In other words : if we have a left representation acting on $v \in V$ like so $\rho^L(g)v$, then, we can construct a right rep with the conjuguate transpose, that will then act on "$v^\dagger$" as such : $v^\dagger \rho^L(g)^\dagger$. Not sure however how to define exactly in what kind of vector space $v^\dagger$ belongs to. $\endgroup$ – Frotaur Dec 9 '17 at 23:58
  • $\begingroup$ @Frotaur: The representation you defined, which I will write as $g \cdot v^\dagger = v^\dagger \rho^L(g)^\dagger$, is in fact a left representation, since $h \cdot (g \cdot v^\dagger) = h \cdot (v^\dagger \rho^L(g)^\dagger) = v^\dagger \rho^L(g)^\dagger \rho_L(h)^\dagger = v^\dagger (\rho^L(h) \rho^L(g))^\dagger$, which is $hg \cdot v^\dagger$. In general, all of these conjugate transposes are really something to do with adjoint operators and the inner product on a Hilbert space - you should have a look at this more. $\endgroup$ – Joppy Dec 10 '17 at 7:45
  • $\begingroup$ Indeed, in my definition this is a right rep, while in yours it is a left representation. However, a problem occured to me in your definition of "right rep" : indeed, it would seem that it is not a representation, since by definition it should be a homomorphism of groups, while your right rep is a antihomomorphism. Maybe it is some extention of the definition of rep, but as far as I know, it does not define a representation as you put it. Let's see if someone has anymore insight ! $\endgroup$ – Frotaur Dec 10 '17 at 15:31

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.