Skip to main content
MathOverflow

Return to Answer

added 121 characters in body
Source Link
user484566
user484566
  • 76
  • 2

$\newcommand\Sym{\mathrm{Sym}}$

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 self-dualorthogonal (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.

$\newcommand\Sym{\mathrm{Sym}}$

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 the odd-dimensional representations are self-dual.

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.

$\newcommand\Sym{\mathrm{Sym}}$

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.

Source Link
user484566
user484566
  • 76
  • 2

$\newcommand\Sym{\mathrm{Sym}}$

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 the odd-dimensional representations are self-dual.

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.