# Beginner's question about representation weights in Lie algebras and their relation to eigenvalues

I am hitting some rough patches in self-learning the representation theory in Lie theory, and specifically, the weight space decomposition. I understand this is not surprising, given the complexity of the topic, but I will not give up based on intimidating nomenclature or definitions.

So I'm asking for an English explanation of what the weights, weight spaces and weight space decomposition are in general based on the particular case below, and explained in here:

In the representation of $${SU}(2)$$ restricted to diagonal matrices $$\begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta} \end{bmatrix}$$

$$R\left( \begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta} \end{bmatrix}\right)=\small \begin{bmatrix}e^{im_1\theta}&&&&\\&e^{-im_2\theta}\\&&\ddots\\&&&e^{im_n\theta} \end{bmatrix}$$

with the integers $$m_1,m_2,\dots,m_n$$ called weights and $$V_i=e^{im_1\theta}$$ called the weight spaces, so that $$V=V_1\oplus V_2\oplus \cdots \oplus V_n$$ is the weight space decomposition in the representation $$R\vert_{\left\{\begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta} \end{bmatrix}\right\}}: \begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta} \end{bmatrix}\to GL(V).$$

In the standard representation $$R:{SU(2)}\to GL(2,\mathbb C)$$ restricted to the diagonal matrices the

• weight spaces are the lines spanned by $$\begin{bmatrix}1\\0 \end{bmatrix}$$ and $$\begin{bmatrix}0\\1 \end{bmatrix}$$ and
• weights are $$1$$ and $$-1.$$

My confusion arises in that in trying to follow up the idea behind this construct I "see" the "weights" as the eigenvalues, in which case, the weights would be $$e^{i\theta}$$ and $$e^{-i\theta}$$. This, together with the eigenvectors, which nicely coincide with the "weight spaces" above would provide a definition I can follow.

But instead of $$e^{i\theta}$$ and $$e^{-i\theta}$$ we are using just $$1$$ and $$-1,$$ which it (naively) feels a bit ad hoc, because we won't always have a diagonalized matrix with exponentials and integers.

Further, is the assumption that $$\begin{bmatrix}1\\0 \end{bmatrix}$$ and $$\begin{bmatrix}0\\1 \end{bmatrix}$$ are eigevectors correct, or are they just standard basis vectors of $$V$$ that coincide in this case?

I presume that the issue is that we not always will go with the standard representation (leaving the matrices as are), but then what is the systematic way of get these weights? Or is it done on a case-by-case basis? What would be other examples?

• "...a weight of a representation is a generalization of the notion of an eigenvalue, and the corresponding eigenspace is called a weight space."
• In the real Lie algebra $${\frak su}(2)$$ has basis given by $$\begin{bmatrix}0 & i\\ i&0\end{bmatrix}, \begin{bmatrix}0 & -1\\ 1&0\end{bmatrix},\begin{bmatrix}i & 0\\ 0&-i\end{bmatrix},$$ while its complexification of $${\frak su}(2,\mathbb C)$$ has basis $$H=\begin{bmatrix}1 & 0\\ 0&-1\end{bmatrix}, X=\begin{bmatrix}0 & 1\\ 0&0\end{bmatrix},Y=\begin{bmatrix}0 & 0\\ 1&0\end{bmatrix},$$ satisfying the commutator relations $$[H,X]=2X, [H,Y]=-2Y, [X,Y]=H.$$

The eigenvalues of $$H$$ are the weights of the representation and if $$v$$ is an eigenvector of $$H$$ with eigenvalue $$\alpha,$$

\begin{align}\small H\cdot (X\cdot v) &= (HX)v= (2X+XH)v=2Xv +XHv=2Xv+X\alpha v= (2+\alpha)X\cdot v\\ \small H\cdot (Y\cdot v) &= (-2+\alpha)Y\cdot v \end{align}

• Initially, and because of the difficulty of handwriting Gothic symbols, I didn't realize the lecture is dealing with the group (not the algebra). I should have realized even intuitively that there is an exponential function there.
• You are mixing up representations of groups and representations of Lie algebras - unless you untangle them none of this will make any sense. $\mathfrak{su}(2)$ doesn't contain those matrices with $e^{i\theta}$ in - those are elements of the group $SU(2)$. Dec 14, 2020 at 19:12
• @MatthewTowers This is a great lead to put all this in order. Thank you! Dec 14, 2020 at 19:28
• @MatthewTowers Before I even have a chance to delve into it, I have to say that it makes perfect sense just thinking about the exponential function involved in going from the algebra to the group! So thanks again! Dec 14, 2020 at 20:04
• I think I got the "players" correctly identified now, but I still don't see the eigenvalues for the group being $1$ and $-1.$ Dec 14, 2020 at 20:29
• the "eigenvalues for the group" aren't $\pm 1$, in fact, it's not clear what "the eigenvalues for the group" means, and getting that straight is the whole problem. The $w$ weight space is the set of vectors on which the element $t_\theta = \begin{pmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{pmatrix}$ acts as multiplication by $e^{i\theta w}$, so in the standard rep, the 1 weight space is spanned by $\begin{pmatrix}1\\0\end{pmatrix}$ and the -1 weight space by $\begin{pmatrix}0\\1\end{pmatrix}$ Dec 14, 2020 at 21:22

The answer consists of two parts: a short 'birds-eye' overview (sections 1-3) and more details and actual definitions to accompany them (sections 4- 6). I once was planning on writing a section 7 but decided I didn't have the time.

There are two things that come together here:

1. Representation theory of abelian groups is much easier than that of non-abelian groups

In fact it is not really different from the theory of eigenvectors and eigenvalues of linear algebra. You just have to be a bit careful in keeping track of the precise relation between the two.

I will edit in later how this plays out in the example of your post, where the Abelian group in question is a subgroup of $$SU(2)$$ isomorphic to $$U(1)$$.

2. If you have a representation $$R$$ of a group $$G$$ on a space $$V$$ and $$A$$ is a subgroup of $$G$$, then $$R$$ is also a representation of $$A$$.

This is trivial if you think about it, but I mention it explicitly because on the surface it might be tempting to think that smaller groups also have smaller representations.

3. How do these two thing work together?

You start with some big non-abelian group $$G$$, in your example $$SU(2)$$. Inside is sitting a smaller abelian subgroup $$T$$, in your example the diagonal matrices. Now when you have an $$n$$-dimensional representation $$R$$ of $$G$$, perhaps irreducible, on a space $$V$$ then we can first view it as a representation of $$T$$. $$T$$ being abelian, everything is much easier and we can understand what this representation looks like. In particular it need no longer be irreducible and we can decompose it as a direct sum of easier to understand subrepresentations (just like in my answer to your earlier question): the weight spaces.

Now this structure of $$V$$ as a direct sum of smaller spaces $$V_i$$ is not necessarily a decomposition into subrepresentations of $$G$$ (unless $$G$$ and $$T$$ were equal from the start) but it might still be useful and simplify computations, especially those involving the Lie algebra.

For instance in your $$\mathfrak{su}(2)$$ example: the original basis can also serve as a basis of the complexification, but we prefer the other one ($$\{X, Y, H\}$$) for practical purposes, because it works really well together with the weight space decomposition of representations. As I said I will add in details later when I find the time.

4. Representations of abelian groups and eigenvalues of matrices

Let's start with your example. We have the group $$T = \left\{ \begin{pmatrix} e^{i \theta} & 0 \\ 0 & e^{- i \theta} \end{pmatrix} \colon \theta \in \mathbb{R} \right\}$$ consisting of all diagonal matrices in the bigger group $$G = SU(2)$$. We'll ignore $$G$$ for now and look at the representation $$R \colon \, T \to GL(\mathbb{C}^{4})$$ given by $$R(\begin{pmatrix} e^{i \theta} & 0 \\ 0 & e^{- i \theta} \end{pmatrix}) = \begin{pmatrix} e^{i m_1 \theta} & 0 & 0 & 0 \\ 0 & e^{i m_2 \theta} & 0 & 0 \\ 0 & 0 & e^{ i m_3 \theta} & 0 \\ 0 & 0 & 0 & e^{ i m_4 \theta} \end{pmatrix}$$ from your post.

Let's focus on two different elements of $$T$$: $$g = \begin{pmatrix} e^{i \theta} & 0 \\ 0 & e^{- i \theta} \end{pmatrix}$$ and $$h = \begin{pmatrix} e^{i \phi} & 0 \\ 0 & e^{- i \phi} \end{pmatrix}$$.

(Of course purely by notation alone we might have $$g = h$$ but for educational purposes we asssume that $$g \neq h$$. Things like $$h = g^3$$ are no problem.)

Now as you already noticed, the vector $$\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ is an eigenvector for the linear operator $$R(g)$$ with eigenvalue $$e^{i m_1 \theta}$$. Similarly $$\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$$ is an eigenvector for $$R(g)$$ with eigenvalue $$e^{i m_2 \theta}$$ etc. Also, when it happens that $$m_1 = m_2$$ we have even more eigenvectors: in that case every vector $$\begin{pmatrix} a \\ b \\ 0 \\ 0 \end{pmatrix}$$ is an eigenvector for $$R(g)$$ with eigenvalue $$e^{i m_1 \theta} = e^{i m_2 \theta}$$.

Now what I wanted to point out to you is what happens if instead of computing the eigenvalues and eigenvectors of the operator $$R(g)$$ we do the same for the operator $$R(h)$$. Please do it yourself if you haven't already.

What we see is:

The eigenvectors (or more generally: eigenspaces) are the same, but the eigenvalues are different.

Of course the first part is the miracle here: a priori we expect everything to be different from one group element to the next. However it is also good to be aware of the different eigenvalues as it indicates that not everything is exactly as easy as we would hope.

So what is going on here? There is a famous theorem in linear algebra stating:

Theorem Let $$S$$ be a set of commuting linear operators on the same space $$V$$. Then there is a decomposition $$V = V_1 \oplus \ldots \oplus V_n$$ such that each $$V_n$$ is a generalized eigenspace for every linear operator $$L$$ from the set $$S$$. (Though possible at different eigenvalues for different $$L$$)

Since $$T$$ is commutative, all the linear operators $$R(g), R(h)$$ etc commute and so we can take $$R(T)$$ to be the set $$S$$ from the theorem. In fact in the example we see that we are even a bit better off and are dealing with actual eigenspaces rather than generalized ones, this is a result of our $$T$$ being compact in addition to abelian. I will come back to that issue later.

So in short: what are the weight spaces? They are the joint eigenspaces for the operators $$R(T)$$.

We also see that answering the question: what are the corresponding weights for each weight space? is slightly more tricky since each $$R(g)$$ can act by a different eigenvalue on a given weight space. I'll write a separate chapter about that.

5. What are the weights?

So how to think about the eigenvalues? If we fix an eigenvector ('weight vector') $$v$$ then apparently there is some map $$\chi \colon T \to \mathbb{C}$$ that assigns to every group element $$g$$ the eigenvalue by which the operator $$R(g)$$ acts on the eigenvector $$v$$. Now let's see if we can understand this map.

It cannot be completely random since $$R$$ is still a representation. If $$R(g)$$ acts by eigenvalue $$a$$ and $$R(h)$$ by eigenvalue $$b$$ then $$R(g)R(h)$$ acts by eigenvalue $$ab$$ by the definition of eigenvalue. But since $$R$$ is a representation this means that group element $$gh$$ also has to act by eigenvalue $$ab$$. In other words $$\chi(gh) = \chi(g)\chi(h)$$.

Since every group element has an inverse this last equation implies that $$\chi$$ cannot take the value 0 and so takes values in $$\mathbb{C}^* = GL(1, \mathbb{C})$$, and more importantly the last equation states that the map $$\chi: T \to GL(1, \mathbb{C})$$ is a group homomorphism. Now we have a word for group homomorphisms into $$GL(1, \mathbb{C})$$: one-dimensional representations!

So $$\chi$$ is a one-dimensional representation of $$T$$ (one-dimensional representations of groups are sometimes called characters, that is why I chose the name $$\chi$$) and perhaps this is not so surprising: it is just the representation $$R$$ where each $$R(g)$$ is restricted to the one-dimensional subspace $$\mathbb{C}v$$.

But... whenever we have a group representation we also have a Lie algebra representation. To every character $$\chi$$ obtained in the way above there is a corresponding Lie algebra representation $$\chi' \colon \mathfrak{t} \to \mathfrak{gl}(1, \mathbb{C}) = \mathbb{C}$$, where $$\mathfrak{t}$$ is the Lie algebra of $$T$$.

This Lie algebra representation is the weight for the representation $$R$$ of the group $$T$$ corresponding to the weight space $$\mathbb{C}v$$.

6. Some notational issues (including: why are weights for $$SU(2)$$ integers?)

• One-dimenisonal Lie algebra representations of abelian groups are only ever called weights if they appear in the context discussed in chapter 2, so if the representation $$R$$ of $$T$$ that we started with was the restriction to $$T$$ of a representation $$R$$ of a bigger, non-abelian group $$G$$.
• When there is talk about 'a weight of $$G$$ for the abelian subgroup $$T$$' without mentioning the specific representation, it means that it is the weight of some finite dimensional representation of $$G$$.
• If also the group $$T$$ is omitted from mentioning and people just talk about 'a weight of $$G$$' (or a weight of a specific representation of $$G$$, but without mentioning which Abelian subgroup $$T$$ we are working with) then the tacit assumption is that $$T$$ is a maximal abelian subgroup.
• The term 'One-dimensional Lie algebra representation' is incredibly pretentious in case the Lie algebra $$\mathfrak{t}$$ of which we take the representation is abelian: in that case the Lie bracket sends every 2 elements to zero and hence $$\mathfrak{t}$$ contains no more information than its underlying vector space. As a result the more common term for 'one-dimensional Lie algebra representation of $$\mathfrak{t}$$' is 'element of the linear dual $$\mathfrak{t}^*$$ of $$\mathfrak{t}$$'. Hence every weight (in the sense of the 3rd bullet point) of $$G$$ is an element of $$\mathfrak{t}^*$$. The converse is not true: the set of all weights (of all finite dimensional reps of $$G$$ for a fixed $$T$$) is an infinite but still only discrete and hence tiny subset of the vectorspace $$\mathfrak{t}^*$$.
• In case $$T$$ and hence $$\mathfrak{t}$$ and hence $$\mathfrak{t}^*$$ are one dimensional (as is the case for $$G = SU(2)$$) we can identify $$\mathfrak{t}^*$$ with $$\mathbb{C}$$ and hence look at elements of $$\mathfrak{t}^*$$ (including all weights) as numbers.

Why in this case these numbers are moreover integers can be seen when going back up one level to groups instead of Lie algebras: each $$\chi$$ ('the exponent of the weight') maps the circle group $$T$$ into $$\mathbb{C}^*$$ by definition, but in practice more specifically into the unit circle inside $$\mathbb{C}^*$$. (Because of compactness.) So... each character $$\chi$$ (corresponding to a weight) is a group homomorphism from the circle group to itself and it is easy to see that the only maps of that type are all of the form $$g \mapsto g^n$$ for some fixed integer $$n$$.

• Wow! This is a great post +1 Jun 23 at 11:14
• Thanks! I am glad people are still reading it after two years! Jun 23 at 11:56