For a complex simple Lie algebra $\frak{g}$, which of its finite dimensional irreducible representations give non-faithful representations of the corresponding simply-connected compact Lie group.

More specifically, can somebody point me to a table detailing, for each series, those dominant weights for faithfulness fails.

Edit: To make my question clearer, I am asking in relation to the answer of this question, which asks when can you build up all representations from the fundamental and antifundamental ones? It is answered that, for an irreducible Lie algebra representation $V$, when the corresponding representation of $G$ is faithful, any other irreducible Lie algebra can be found in a tensor product $V^{\otimes k}$, for sufficiently high $k$. The answer is qualified with the following comment:

One mild warning: there is an obvious representation of $\frak{𝔰𝔬}(𝑛)$ which is not a faithful representation of the corresponding simply-connected compact Lie group when $n\geq 3$

Is the "obvious representation" of $\frak{so}(n)$ the only representation for which this happens, or are there others?

  • $\begingroup$ AFAIK, all finite dimensional irreducible representations of $\mathfrak{g}$ are integrable. So, no need to worry about the difference between $\mathfrak{g}$ and $G$. AFAIK, for a integral positive weigth $\omega$, it will not give rise to a faithful representation if $(\omega,\beta^\vee)=0$ for a simple root $\beta$ (which seems equivalent to me). $\endgroup$ – user66288 Apr 16 at 12:19
  • $\begingroup$ these are exactly the irreducible representations with highest weight (which is a character on a maximal torus $T$ of the simply connected compact group $G$), which is trivial restricted to the centre of G . $\endgroup$ – Venkataramana Apr 16 at 13:17
  • $\begingroup$ @Venkataramana: What does this look like dually on the Lie algebra side? $\endgroup$ – Nadia SUSY Apr 16 at 13:19
  • 1
    $\begingroup$ In order to get a faithful representation of the spinor group, you must use a spinor representation somehow. So contrary to the impression you got from that comment, the standard representation is not atypical when viewed among all simple modules. What it is true that you can build all non-spinor fundamental weights by tensor products and decomposing. $\endgroup$ – Victor Protsak Apr 16 at 15:17

Let $G_{sc}$ be as in the answer by Victor Protsak and let $\varpi_1$, $\ldots$, $\varpi_l$ be the fundamental dominant weights.

Let $\lambda$ be a dominant weight and write $\lambda = \sum_{i = 1}^l a_i \varpi_i$ for $a_i \in \mathbb{Z}_{\geq 0}$.

Then the irreducible representation of $G_{sc}$ with highest weight $\lambda$ is faithful precisely in the following cases:

  • Type $A_{l}$ ($l \geq 1$): $\gcd(l+1, a_1+2a_2+\cdots+la_l) = 1$.
  • Type $B_l$ ($l \geq 2$): $a_l$ is odd.
  • Type $C_l$ ($l \geq 2$): $a_1 + a_3 + a_5 + \cdots$ is odd.
  • Type $D_l$ ($l \geq 4$): $l$ is odd and $a_{l-1} + a_l$ is odd
  • Type $G_2$: always
  • Type $F_4$: always
  • Type $E_6$: $a_1 - a_3 + a_5 - a_6$ is not divisible by $3$.
  • Type $E_7$: $a_2 + a_5 + a_7$ is odd
  • Type $E_8$: always

This can be determined by a direct computation. The kernel of any irreducible representation of $G_{sc}$ lies in $Z(G_{sc})$, which is finite. Furthermore, you can describe $Z(G_{sc})$ explicitly and compute the action of any $z \in Z(G_{sc})$ in an irreducible representation (it is of course always multiplication by some scalar). See for example Chapter 3, Lemma 28 in "Lectures on Chevalley Groups" by Steinberg.

Note above that in types $G_2$, $F_4$ and $E_8$ the center of $Z(G_{sc})$ is trivial so every irreducible representation is faithful.

Also, for type $D_{2l}$ the center is not cyclic so no irreducible representation is faithful.

  • $\begingroup$ Great! Just what I was hoping for. Thanks a lot! $\endgroup$ – Nadia SUSY Apr 16 at 16:21
  • $\begingroup$ sorry, but should the $m_i$'s in the exceptional cases be $a_i$'s? $\endgroup$ – Nadia SUSY Apr 16 at 16:23
  • $\begingroup$ @NadiaSUSY: Right, thanks. $\endgroup$ – spin Apr 16 at 19:41

By the highest weight theory, finite-dimensional irreducible representations of a finite-dimensional simple Lie algebra $\mathfrak{g}$ are parametrized by the dominant weights $\lambda$ in the weight lattice $Q$. On the other hand, the analogous representations of the adjoint form $G_{\rm ad}$ of the corresponding simple Lie group are parametrized by the dominant weights in the root lattice $P$. The difference is measured by the finite abelian group $Q/P$, which may be identifed with the dual of the center $Z$ of the simply-connected form $G_{\rm sc}$ (in fact, $Z\simeq\! P^\vee/Q^\vee$). A representation of $G_{\rm sc}$ is faithful if and only if its restriction to $Z$ is faithful if and only if $\lambda$ is not contained in any proper sublattice of $Q$.

This explained and tabulated in terms of the corresponding root systems in nearly all introductory Lie theory textbooks, but I am partial to exposition in Goto and Grosshans, where I first learned it.

  • $\begingroup$ I wasn't able to type $Z\simeq P^\vee/Q^\vee$ in the answer at first due to a weird glitch. $\endgroup$ – Victor Protsak Apr 16 at 15:20

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