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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.

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](http://math.stackexchange.com/questions/631935/cartan-matrix-for-an-exotic-type-of-lie-algebra)2014-01-10

1 Answers 1

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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?2012-01-21
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    (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)..."2012-01-21
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    (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.2012-01-21
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    @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 $\alpha_i$.2012-01-22
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    @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 :)2012-01-22
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    Thanks for the reply. I did mean that you were choosing $\{ E_{i i+1} \}_{i=1} ^{i=n-1}$ as a basis for each of the simple-root spaces. About the later ones, it would be equally great if you can tell me a specific reference where some explicit example would be worked out which illustrates the issues I asked about in comment (3) and (4). Its hard to filter out this kind of specific example calculation from say the tomes of Fulton and Harris or Knapp - though they are great books for the general concept.2012-01-22
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    @Anirbit sorry but I can't think of any references which work out specific examples of this stuff in detail. James Humphrey's "Introduction to Lie Algebras" is the standard reference for this material (but it's quite terse). Samelson's "Notes on Lie Algebras" is easier to read (but hard to find). Actually, Erdmann and Wildon's "Introduction to Lie Algebras" might be a good place to look. It's supposed to be an undergraduate text and contains details many other books leave out. I read it a number of years ago and thought it was very easy to read.2012-01-23
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    Thanks for the additional edit. I have two questions about that (1) I am aware that under the adjoint action of the lie algebra on itself it undergoes a decomposition like $\cal{g} = t \oplus _\alpha \cal{g}_\alpha$ You seem to be claiming that $V$, any representation of $\cal{g}$ has a similar decomposition of the kind $V = \oplus _{\alpha \in t^*} V_\alpha$. Can you explain as to why this should exist?2012-01-28
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    (2) You seem to suggest that the "highest weight" is a root such that to which if any simple root is added then the sum is not a weight anymore. I earlier had the idea that a vector is called the "highest weight" if it is annihilated by all "positive" root spaces. How do these match up?2012-01-28
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    I guess you are shifting between two different pictures of "weights" - once as representations of the Cartan subalgebra and once as elements of the dual of the Cartan subalgebra. Also this is different from the notion of "weights of a representation" where one seems to be mean those specific weights which appear in the decomposition of the representation into irreducible modules (each labelled by a weight.) Am I right in this understanding?2012-01-28
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    Dear all, perhaps you experts have interests to solve this problem: [cartan-matrix-for-an-exotic-type-of-lie-algebra](http://math.stackexchange.com/questions/631935/cartan-matrix-for-an-exotic-type-of-lie-algebra)2014-01-10