$E\_6$'s Dynkin diagram is a line of 5 vertices, which we will number 1...5, and a sixth one attached to #3, which we will ignore.


$\dim V\_{\omega\_2} = 351 = \dim V\_{2\ \omega\_1}$, where $\omega\_i$ denotes the corresponding fundamental weight and $V$ the irrep with that high weight. (So $\dim V\_{\omega\_1} = 27$, the number of lines on a cubic surface.)


"Why" do these two irreps have the same dimension? I'm not sure what I'm asking, obviously, but here's an attempt: do they become isomorphic when restricted to some large isomorphic subgroups of $E\_6$?


Incidentally, $V\_{\omega\_2} = \wedge^2 V\_{\omega\_1}$, and $Sym^2 V\_{\omega\_1} = V\_{2\omega\_1} \oplus (V\_{\omega\_1})^\*$. So maybe the best answer is "the $Sym^2$, which is always larger than the $\wedge^2$, is in this case only larger by something of the same dimension as the original space ($V\_{\omega\_1}$)".



Why do these two irreps of $E_6$ have the same dimension?
$\newcommand\Sym{\mathrm{Sym}}$


An extended comment which more or less suggests that your suggested answer might be as good as one can do.


If $G$ has a representation on $V$ which preserves a symmetric trilinear form on $V$, then $\Sym^3(V)$ has a $G$-invariant vector and thus $\Sym^2(V) \otimes V$ has a $G$-invariant vector and thus $V^{\vee}$ is a constituent of $\Sym^2(V)$. So for any such $G$ the virtual representations


$$\wedge^2(V), \quad [\Sym^2(V)] - [V^{\vee}]$$


are both actual representations and both have the same dimension. The Dickson invariant of $E\_6$ acting on the $27$-dimensional representation $V$ is the corresponding form in this case.


To give a related example, if you take $V$ and restrict to $F\_4$ then it decomposes as $U \oplus \mathbf{C}$. The action of $F\_4$ on $U$ also admits an invariant cubic form. Hence you obtain a pair of corresponding representations of $F\_4$ of dimensions $325$ which are not isomorphic. Unlike the case of $E\_6$, however, neither of these are irreducible. This is clear in one case, because the action of $F\_4$ on $U$ preserves a quadratic form. So now we have $[U^{\vee}] = [U]$ and decompositions


$$[\Sym^2(U)] - [U] = 1 + 324,$$
$$[\wedge^2(U)] = 273 + 52,$$


where the numbers refer to irreducible representations of the corresponding dimension.


You also see from this that the restrictions of your $351$ dimensional representations to $F\_4$ are still different. But they are still different even when you restrict to the principal $\mathrm{SL}\_2$. The $27$-dimensional representation $V$ restricts to the principal $\mathrm{SL}\_2$ as a sum of representations $U\_1 \oplus U\_7 \oplus U\_{19}$ where a representation of $\mathrm{SL}\_2$ is determined by its dimension. But now:


$$\wedge^2(V) = \wedge^2(U\_1 + U\_7 + U\_{19}) = \wedge^2(U\_7) + \wedge^2(U\_{19}) + U\_7 + U\_{19}+ U\_7 \otimes U\_{19},$$


from which we see that the $351$-dimensional representation $\wedge^2(V)$ has no $\mathrm{SL}\_2$-invariants because odd dimensional irreducible representations don't admit symplectic forms.
On the other hand,


$$\Sym^2(V) - [V^{\vee}] = \Sym^2(U\_1 + U\_7 + U\_{19}) - (U\_1 + U\_7 + U\_{19})
= \Sym^2(U\_1) + \Sym^2(U\_7) + \Sym^2(U\_{19}) - U\_1 + \ldots $$


has at least a $2$-dimensional space of invariants because all representations are self-dual and thus the the odd-dimensional representations are orthogonal (of course one can see they are orthogonal more directly by constrution).


So I think the conclusion is that the existence of an invariant symmetric cubic form guarantees the existence of two representations of the same dimension $\binom{n}{2}$ which have no reason to be related, and in the case of $E\_6$ they just both happen to be irreducible.


