Reps of groups and reps of algebras

Question

I've got what might be a couple of very basic questions on the fundamental representations of locally isomorphic semi-simple Lie groups and their relationship to representations of the corresponding Lie algebra. I know (from reading Cornwell, 'Group Theory in Physics') that every representation of a semi-simple Lie algebra exponentiates to give a representation of the unique, simply connected semi-simple Lie group associated with that algebra. (So for example, every representation of the algebra su(2) exponentiates to give a representation of the group SU(2).) I also know that, at least in some cases (such as with the group SO(3), which also has su(2) as its algebra), it is not the case that every representation of the algebra exponentiates to give a representation of a given non-simply-connected group associated with that algebra. The first non-trivial representation of SO(3), for example, is not obtainable by exponentiating the fundamental (2-dimensional) representation of su(2) (with highest weight 1/2), but is rather obtained by exponentiating the 3-dimensional representation of the algebra (with highest weight 1).

So I have two questions on the same theme as the above.

1. Is it always the case that the fundamental representation(s) of the various non-simply-connected semi-simple Lie groups associated with a given semi-simple Lie algebra are not obtainable by exponentiating the fundamental representation(s) of the algebra?

2. Do the fundamental representation(s) of each of the different non-simply-connected groups correspond to different representations of the algebra? (That is, will two groups that are locally but not globally isomorphic always have different fundamental representations?)

Any knowledge anyone can bring to bear on this would be really appreciated. (I'm pretty sure the second in particular is trivial, but I'm not sure as at the level I'm at in physics we pretty much always work with representations of the algebras and I'm not familiar with how they relate to the representations of the groups.) Thanks a lot.

Answer 1

Let us consider for simplicity the complex semi-simple case. As you mention, every irreducible representation $\rho$ of a semi-simple finite dimensional Lie algebra $g$ integrates to a representation of the corresponding simply connected Lie group $G$. But even if we suppose $\ker\rho=0$, the representation of $G$ may well have a kernel, a finite central subgroup $H$ of $G$. So we obtain representations of $G/H'$ where $H'$ is a finite central subgroup contained in $H$, but not of other groups with Lie algebra $g$.

So one could ask: given a Lie group $G'$ with Lie algebra $g$, when does a representation $\rho$ of $g$ with highest weight $\Lambda$ integrates to a representation of $G'$? The answer: if and only if $\Lambda$ belongs to the character lattice of a maximal torus $T$ of $G$. (Recall that the character lattice of $T$ is the discrete additive subgroup of the dual of the Lie algebra of $T$ spanned by the differentials of the homomorphisms $T\to\mathbf{C}^{\ast}$.)

One can also ask a similar question: given a representation $\rho$ of $g$ with $\ker\rho =0$ and highest weight $\Lambda$, how to compute the kernel of the resulting representation $R$ of the simply connected group $G$? The answer is as follows. The dual of the Lie algebra $t$ of $T$ contains the weight lattice $P$ and the root lattice $Q$. Recall that the weight lattice is the lattice spanned by the fundamental weights and the root lattice is spanned by the roots i.e. the weights of the adjoint representation. We have $Q\subset P$. Consider the dual lattices $Q^\vee\subset P^\vee$; these are formed by all elements $a$ of the Lie algebra of $T$ such that $l(a)\in\mathbf{Z}$ for all $l\in P$, resp. all $l\in Q$.

Inside $P^\vee$ there is a sublattice formed by all $a\in t$ such that $\Lambda(a)\in\mathbf{Z}$; it contains $Q^\vee$ and the quotient by $Q^\vee$ is naturally identified with the kernel of $R$ (by exponentiating). There is a similar result when $\rho$ is reducible -- in that case we just consider the sublattice formed by all elements of the tangent space such that all weights that take integral values on them.

Example: when $G=SL_2(\mathbf{C})$, we can identify $Q$ with the sublattice of $\mathbf{R}$ spanned by 1 and $R$ will be spanned by $\frac{1}{2}$. Then the representations with integral weight will have kernel $\mathbf{Z}/2$ and the representations with half-integral weight will have trivial kernel.

Hopefully one of these two questions also covers the questions of the posting.

Answer 2

To supplement algori's answer: The first interesting case (for the Lie algebra of type $A_1$ over $\mathbb{C}$) is already noted in the question: the odd dimensional irreducible representations exponentiate to both the simply connected and the adjoint group, but not the even dimensional ones including the single fundamental representation. In type $B_2$ there are two fundamental representations, of dimensions 4 and 5. The latter one gives the standard realization of the adjoint group as a special orthogonal group in 5 variables, while the former one is the "spin" representation of the simply connected covering group but is not a representation of the adjoint group. As algori notes, which fundamental representations exponentiate to which group depends on the position of its highest weight in the weight lattice of a maximal torus of the group relative to the root lattice.

However, in all cases basic Lie theory guarantees that representations belonging to different highest weights (including the fundamental ones) are non-isomorphic.

In the case of real semisimple Lie algebras and groups, similar but more complicated things happen as seen in a number of math and physics textbooks.