# There are no irreducible representations of dimension less than $7$ for the Lie algebra $\mathfrak g_2$.

I know that any finite dimensional irreducible representation of $$\mathfrak g_2$$ must appear in some tensor power of the standard representation $$V$$ which is seven dimensional. But still there may be $$3$$-dimensional irreducible representation $$W$$ included in $$V\otimes V$$ which is not a subrepresentation of $$V$$. Can we prove that this is not possible without using the Weyl character formula?

• The same question had been asked earlier. Looks like the idea I used appeared then also, but only in a comment. That one has no answers, so cannot serve as duplicate target. As I happened to answer this, I should not vote, but anything the others want is fine with me also. Including relocating the answers to the other thread. Jul 27, 2022 at 13:23
• Here is how it would go using Weyl's dimension formula. Jul 27, 2022 at 14:23

One possibility is to use the fact that the formal character of an irreducible representation is invariant under the action of the Weyl group. In the case of $$\mathfrak{g}_2$$ the Weyl group has order twelve, and the stabilizer of a non-zero weight has order at most two. Either by inspection, or due to the fact that the stabilizer of a weight in the closure of the dominant Weyl chamber is generated by the simple reflections in it.

This already forces the dimension of a non-trivial representation to be at least $$12/2=6$$. Furthermore, for $$\mathfrak{g}_2$$ the weight lattice coincides with the root lattice, so $$\mathfrak{sl}_2$$-theory implies that the zero weight is automatically also a weight of any non-trivial representation. The minimum dimension thus goes up to $$6+1=7$$.

• Very nice argument. Jul 27, 2022 at 15:05
• This is very nice. Only part I did not get is the ''order of the stabilizer must be at most two''. I should think about the stabilizer and orbits then. Jul 28, 2022 at 13:10
• @MehmetK Stabilizers and orbits, indeed. Another way of looking at is to recall that the Weyl group is the dihedral group of order twelve. There are six rotations in it, and no non-zero plane vector is stable under a rotation. So the orbit of any non-zero weight has size at least six. Jul 28, 2022 at 21:11
• Ooh, I see. Thank you very much sir. Jul 29, 2022 at 18:14

If you are willing to use that $$\mathfrak g_2$$ is simple, then you can use the fact that any non-zero representation is injective and that $$\mathfrak g_2$$ has dimension $$14$$. So for a representation in dimension $$m$$ you need $$m^2-1\geq 14$$, so any non-trivial representation of $$\mathfrak g_2$$ has to have dimension at least $$4$$. You can also rule out dimension $$4$$ easily, since $$\mathfrak{sl}(4,\mathbb R)$$ has no simple subalgebras of codimension $$1$$.

• This is very helpful, thank you. But in my question, 3 was actually a random number less than 7. Can we say something looking at weight lattice of $\mathfrak{g}_2$, Weyl group or the fundamental weights specifically. Jul 26, 2022 at 21:55
• You could probably cook up an elementary argument that the lowest possible dimension has to be attained by a fundamental representation, but I am not sure about that. Jul 27, 2022 at 6:44
• This is a good start. It is not entirely obvious to me how to best close the gap, but +1 obviously. My suggestion assumes a bit extra knowledge. Jul 27, 2022 at 12:48