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I am a bit confused about the notion of "fundamental weights".

In a complexified setting, I am thinking of my Lie algebra to be decomposed as, $\cal{g} = \cal{t} \oplus _\alpha \cal{g}_\alpha$ where the $\cal{g}_\alpha$ are the root-spaces. Now given a root $\alpha_j$, one defines its co-root $H_{\alpha_j} \in [\cal{g}_{\alpha _j}, \cal{g}_{-\alpha _j}]$ such that $\alpha_j (H_{\alpha _j}) = 2$

  • Now one seems to define the "fundamental weights" as a set rank $G$ elements $\omega_i \in t^*$ such that, $\omega_i (H_{\alpha _j}) = \delta_{ij}$

    • In the above definition is it necessary that the $\alpha_j$ have to be simple roots? (..i get this feeling when looking at examples..) I guess one can get away by defining the action of the fundamental weights on the co-roots of simple roots only because the co-roots are themselves enough to give a basis for $t^*$ just like the simple-roots. Is that right?

    • For the case of $SU(n)$ one chooses the simple root spaces to be the spans of the matrices $E_{ij}$ - which have a $1$ at the $(i,j)$ position and a $0$ everywhere else. If the Cartan subalgebra is spanned by matrices of the form $H_\lambda = diag(\lambda_i)$, then one has the roots $\alpha_{ij}$ defined as, $[H_\lambda,E_{ij}] = \alpha_{ij}(H_\lambda)E_{ij} = (\lambda_i - \lambda_j)E_{ij}$ Now since $\alpha_{ji} = - \alpha_{ij}$, one would search for the co-root $H_{\alpha_{ij}} \in [E_{ij},E_{ji}]$. Hence I would have naively expected that $H_{\alpha_{ij}} = E_{ii} - E_{jj}$ for all pairs of $i<j$.

But why is it that in literature I see the co-roots of $SU(N)$ to be taken as, $H_{\alpha _ {i i+1}} = E_{ii} - E_{i+1,i+1}$? Is this again a question of some standard choice of basis?

  • From the above how does it follow that the fundamental weights $\omega_i$ of $SU(N)$ are given as $\omega_i (H_\lambda) = \sum _{k=1} ^{k=i} \lambda_k$ ?

  • How is all the above related to the idea that there are $N-1$ fundamental representations of $SU(N)$? And how are they demarcated?

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Dear Anirbit, perhaps you have interests to illuminate this problem: cartan-matrix-for-an-exotic-type-of-lie-algebra –  mystery Jan 10 at 20:25

1 Answer 1

up vote 7 down vote accepted

Fundamental weights correspond to fundamental roots (i.e. simple roots). Each choice of simple roots leads to a different choice of fundamental weights. There aren't really any fundamental weights associated with other (non-simple) roots (or at least this terminology isn't standard to my knowledge).

[Note: The rank of $\mathfrak{sl}_N$ (or equivalently $SU(N)$) is $N-1$. I will set $\ell=N-1$.]

Basics: First, a set of simple roots must be chosen (any two systems of simple roots are conjugate under the action of the Weyl group). Say $\{\alpha_1,\dots,\alpha_\ell \}$ is you set of simple roots. Suppose we have also fixed a set of Chevalley generators $\{ E_i, F_i, H_i \;|\; i=1,\dots,\ell \}$ so these are elements such that $H_i \in [\mathfrak{g}_{\alpha_i},\mathfrak{g}_{-\alpha_i}]$ such that $\alpha_i(H_i)=2$ and $[E_i,F_i]=H_i$ where $E_i \in\mathfrak{g}_{\alpha_i}$ and $F_i \in\mathfrak{g}_{-\alpha_i}$. Then $\alpha_j(H_i)=a_{ji}$ = the $i,j$-entry of the Cartan matrix (or the $j,i$-entry of the Cartan matrix, depending on whose convention you are using) so in particular $\alpha_i(H_i)=a_{ii}=2$.

Next, what you have for the fundamental weights is not quite correct. The fundamental weights $\{\omega_1,\dots,\omega_\ell \}$ form a basis for $t^*$ which is dual to the (basis of) simple coroots $\{H_1,\dots,H_\ell\}$. In other words, $\omega_i(H_j)=\delta_{ij}$ (the Kronecker delta: $\delta_{ii}=1$ and $\delta_{ij}=0$ for $i\not=j$). In particular, $\omega_i(H_i)=1$ (not $2$).

Next, take a finite dimensional irreducible $\mathfrak{g}$-module. From the theory we know it is a highest weight module, say $V(\lambda)$ which is the direct sum of weight spaces. These weights are of the form $c_1\omega_1+\cdots+c_\ell\omega_\ell$ where $c_i \in \mathbb{Z}$ (integral linear combinations of fundamental weights). In particular, the roots of $\mathfrak{g}$ along with $0$ (the zero functional) are the weights of the adjoint representation. So roots are integral linear combinations of fundamental weights. Actually, it turns out that $\alpha_i = a_{i1}\omega_1+a_{i2}\omega_2+\cdots+a_{i\ell}\omega_{\ell}$ so the Cartan matrix (or its transpose) is the change of basis matrix from fundamental weights to simple roots. The importance of the fundamental weights is that they form a basis for the lattice of weights of finite dimensional representations of $\mathfrak{g}$.

So $\{H_1,\dots,H_\ell\}$ (simple co-roots) form a basis for $t$. Both $\{\alpha_1,\dots,\alpha_\ell\}$ (simple roots) and $\{\omega_1,\dots,\omega_\ell\}$ (fundamental weights) are bases for $t^*$. The fundamental weight basis is dual to the simple co-root basis. And the Cartan matrix is a change of basis matrix from the simple roots to the fundamental weights.

Next, for $\mathfrak{sl}_N$ (the root space decomposition is for the Lie algebra not the Lie group $SU(N)$). While $E_{ij}$ ($i \not= j$) are root vectors, only $E_{i,i+1}$ and $E_{i+1,i}$ are in simple root spaces. In particular, $E_i = E_{i,i+1} \in (\mathfrak{sl}_n)_{\alpha_i}$ (the $\alpha_i$ root space) and $F_i = E_{i+1,i} \in (\mathfrak{sl}_n)_{-\alpha_i}$ (the $-\alpha_i$ root space). Then $H_i = [E_i,F_i] = E_{i,i+1}E_{i+1,i} - E_{i+1,i}E_{i,i+1} = E_{i,i} - E_{i+1,i+1}$ (the simple co-roots). Your other $E_{ii}-E_{jj}$ are co-roots as well just not necessarily simple co-roots.

If $H_\lambda = \mathrm{diag}(\lambda_1,\dots,\lambda_\ell)$, then $H_\lambda=\lambda_1H_1+(\lambda_1+\lambda_2)H_2+\cdots+(\lambda_1+\cdots+\lambda_\ell)H_\ell$. For example: Consider $H_\lambda = \mathrm{diag}(\lambda_1,\lambda_2,\lambda_3)$. Keep in mind that since $H_\lambda \in \mathfrak{sl}_3$ it has trace=0, so $\lambda_3=-\lambda_1-\lambda_2$. Thus $$ \begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix} = \begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & -\lambda_1 & 0 \\ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & \lambda_1+\lambda_2 & 0 \\ 0 & 0 & -\lambda_1-\lambda_2 \end{bmatrix} $$

$$= \lambda_1\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}+(\lambda_1+\lambda_2)\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$

So in general, $\omega_i(H_\lambda) = \omega_i(\lambda_1H_1+(\lambda_1+\lambda_2)H_2+\cdots+(\lambda_1+\cdots+\lambda_\ell)H_\ell) = \lambda_1+\cdots+\lambda_i$ since $\omega_i(H_i)=1$ and $\omega_i(H_j)=0$ for $i \not= j$.

The $N-1$ fundamental representations of $SU(N)$ are the highest weight representations with highest weights $\omega_1,\dots,\omega_{\ell}$. These are often denoted $V(\omega_1),\dots,V(\omega_\ell)$. All other (finite dimensional) irreducible representations appear as subrepresentations of tensor products of these representations.

Edit: I will try to add a brief account highest weight modules. Here goes...

Let $\mathfrak{g}$ be a finite dimensional semi-simple Lie algebra. Then every finite dimensional $\mathfrak{g}$-module (i.e. representation) is completely reducible (can be written as a finite direct sum of irreducible modules). Then it can be shown that each irreducible module is a highest weight module. So in the end, if we know everything about highest weight modules, then we'll essentially know everything about all modules.

What is a highest weight module? Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra with Cartan subalgebra $\mathfrak{h}$ (Cartan subalgebra = maximal toral subalgebra = your "$t$"). In addition fix a set of simple roots $\{ \alpha_1,\dots,\alpha_\ell\}$ and fundamental weights $\{ \omega_1,\dots,\omega_\ell \}$.

Let $V$ be a $\mathfrak{g}$-module. Then $V$ is a weight module if $V = \oplus_{\mu \in \mathfrak{h}^*} V_\mu$ (the direct sum of weight spaces) where $V_\mu = \{ v\in V \;|\; h \cdot v = \mu(h)v \}$. If $V_\mu \not= \{0\}$, then $V_\mu$ is a weight space and $\mu \in \mathfrak{h}^*$ is called a weight. [Example: If you consider $\mathfrak{g}$ itself as a $\mathfrak{g}$-module, then the weights of the adjoint action are the roots along with the zero functional.] So if $v \not=0$ is in the $\mu$ weight space and $h \in \mathfrak{h}$, then $v$ is an eigenvector for the action of $h$ with eigenvalue $\mu(h)$. Thus $V_\mu$ is the simultaneous eigenspace for the operators given by the action of each $h \in \mathfrak{h}$ with eigenvalues $\mu(h)$.

It can be shown that a finite dimensional irreducible $\mathfrak{g}$-module is a weight module and there exists a unique weight $\lambda \in \mathfrak{h}^*$ such that $\lambda+\alpha_i$ is not a weight for all $i=1,\dots,\ell$. So thinking of $\alpha_i$ as pointing "up" in some sense, $\lambda$ is as high as you can go. It's the highest weight. Next, every weight in the module is of the form $\lambda-(c_1\alpha_1+\cdots+c_\ell\alpha_\ell)$ for some non-negative integers $c_i$ (all weights lie below the highest weight). Also, the structure of an irreducible module is completely determined by its highest weight. So if $V$ and $W$ are irreducible highest weight modules, then $V \cong W$ if and only if $V$ and $W$ have the same highest weight. Moreover, it turns out you can construct (a unique) irreducible highest weight module for any $\lambda \in \mathfrak{h}^*$. We usually call this module something like $V(\lambda)$. However, it turns out that although $V(\lambda)$ is an irreducible highest weight module, it is finite dimensional if and only if $\lambda=c_1\omega_1+\cdots+c_\ell\omega_\ell$ where each $c_i$ is a non-negative integer.

Fix a set of non-negative integers $c_i$. Then suppose we tensor product the highest weight module $V(\omega_i)$ (a fundamental module) $c_i$-times with itself and then tensor all of these together. Then we will have a (reducible) module which contains a copy of the irreducible highest weight module $V(c_1\omega_1+\cdots+c_\ell\omega_\ell)$. Thus the fundamental modules give us a way of constructing all finite dimensional irreducible highest weight modules [although the tensor product will include copies of other irreducible modules in general so we'll have to filter out this unwanted extra stuff.]

Your final question. Given a highest weight for $SU(N)$ (equivalently $\mathfrak{sl}_N$), how does one write down matrices for the action associated with the corresponding highest weight module? That is a non-trivial, quite complicated computation. Even the answer for $SU(3)$ is complicated. So I'm going to pass on that one. :)

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Thanks a lot for this almost text-book kind of answer! It was awesome. I have corrected some of my typos that you pointed out. I have some further clarifications to ask about what you said - (1) Shouldn't your definition of the Cartan matrix be $\alpha_j (H_i) = a_{ji}$ to be consistent with your convention of saying $\alpha_i = a_{ij}\omega_j$ ? (2) About the simple roots of $SU(N)$, I guess you are choosing them to be the set $\{ E_{i i+1} \}_{i=1} ^{i=n-1}$.. right? –  Anirbit Jan 21 '12 at 20:59
    
(3) I am not very clear about the idea of this "highest weight module". If you could kindly add in a few more lines of explanation like I did not understand what you meant in that line, "..say $V(\lambda)$ which is the direct sum of weight spaces. These weights are of the form $c_1\omega_1+\cdots+c_\ell\omega_\ell$ where $c_i \in \mathbb{Z}$ (integral linear combinations of fundamental weights)..." –  Anirbit Jan 21 '12 at 21:04
    
(4) About the issue of "fundamental representations' of $SU(N)$ I guess I did not make my question very clear. Can you kindly elaborate on as to how does picking a the highest weight say some $\omega_i$ (for $i \in \{ 1,...,n-1\}$) specifies the representation. Like if I pick some $g \in SU(N)$ then how do I write down the matrix for $g$ knowing what the highest weight is - say some $\omega_i$. I know how to do this for $SU(3)$ bcause that can be written in the language of quantum angular momentum but otherwise I don't see anything. –  Anirbit Jan 21 '12 at 21:05
    
@Anirbit Yes. About (1), you are correct. I used one convention one place and another further down :( As for (2), yes and no. $E_{ii+1}$ are elements of simple root spaces. But the simple roots are linear functionals (elements of $t^*$ instead of $\mathfrak{g}$). The $E_{ii+1}$ are root vectors. Root vectors are elements of the algebra whose weights (think of eigenvalues) are roots. So in some sense $E_{i,i+1}$ are basically the eigenvectors to go with the eigenvalues $\alpha_i$. –  Bill Cook Jan 22 '12 at 20:13
    
@Anirbit I don't have time right now, but I'll try to edit the post later to address (3) and (4)...although you'll need a textbook for a real full answer :) –  Bill Cook Jan 22 '12 at 20:13

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