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I am self studing classical Lie algebras and their representations from the book Representation Theory – A First Course by William Fulton and Joe Harris.

In the chapter on representations of symplectic Lie algebras $\mathfrak{sp}_{4}\mathbb{C}$, the four-dimensional vector space $V$ is defined as the standard representation of $\mathfrak{sp}_{4}\mathbb{C}$. They next consider the exterior power $\bigwedge^{2} V$.

I have difficulty understanding the statement “$\bigwedge^{2} V$ cannot be irreducible representation of $\mathfrak{sp}_{4} \mathbb{C}$, since the corresponding group action of Lie group $\mathrm{Sp}_{4} \mathbb{C}$ on $V$ by definition preserves the skew form $Q \in \bigwedge^{2} V$”.

I am not quite comfortable with Lie algebra and Lie group association. Can anybody please explain this to me? Moreover I am not sure why the skew symmetric form $Q$ belongs to $\bigwedge^{2} V$.

  • Regarding the last sentence: in my copy of the book, it states instead that “$Q ∈ ⋀^2 V^* ≅ ⋀^2 V$”. So they claim that $Q$ is an element of $⋀^2 V^$, and then also claim that $⋀^2 V^$ is isomorphic to $⋀^2 V$. – Jendrik Stelzner Dec 22 '22 at 11:46
  • Yes, you are right. That's a mistake from my end. – simu tiyam Dec 22 '22 at 16:23

1 Answers1

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Let us recall some background on the actions of Lie algebras of bilinear forms. Let $$ be a $$-Lie algebra and let $V$ be a representation of $$, where $$ is some field with $\operatorname{char} ≠ 2$.

  • An element $v$ of $V$ is called invariant if $xv = 0$ for all $x ∈ $.

Note that a non-zero invariant element always spans a one-dimensional trivial subrepresentation. An irreducible representation must therefore not contain any non-zero invariant elements, unless it is the one-dimensional trivial representation.

  • A bilinear form $β \colon V × V \to $ is called $$-invariant if $$ β(xv, w) + β(v, xw) = 0 \qquad \text{for all $x ∈ $ and all $v, w ∈ V$.} $$ If $$ is the Lie algebra of a Lie group $G$, then this corresponds (roughly) to $β$ being $G$-invariant in the sense that $$ β(gv, gw) = β(v, w) \qquad \text{for all $g ∈ G$, $v, w ∈ V$.} $$

This notion of “$$-invariant” is a special case of the general notion of “invariant”.

  • The vector space of bilinear maps $\newcommand{\Bil}{\operatorname{Bil}}\Bil^2(V, ℂ)$ becomes again a representation of $$ via $$ (x β)(v, w) = β(xv, w) + β(v, xw) \qquad \text{for all $x ∈ $, $β ∈ \Bil^2(V, )$, $v, w ∈ V$.} $$ A bilinear form $β$ is $$-invariant if and only if it invariant in the general sense.

We have some standard homomorphism/isomorphisms regarding $\Bil^2(V, )$.

  • The usual isomorphism of vector spaces $\Bil^2(V, ) ≅ (V ⊗ V)^*$ is an isomorphism of representations. The usual injective linear map $V^* ⊗ V^* \to (V ⊗ V)^*$ is a homomorphism of representations. If $V$ is finite-dimensional, then it is thus an isomorphism of representations.

We can also restrict our attention to alternating bilinear forms, or symmetric bilinear forms.

  • Let $\newcommand{\Alt}{\operatorname{Alt}} \Alt^2(V, )$ be the linear subspace of $\Bil^2(V, )$ consisting of alternating bilinear forms. This is $$-subrepresentation of $\Bil^2(V, )$. The usual isomorphism of vector spaces $\Alt^2(V, ) ≅ (⋀^2 V)^*$ is an isomorphism of representations. The usual injective linear map $⋀^2 (V^*) \to (⋀^2 V)^*$ is a homomorphism of representations. If $V$ is finite-dimensional, then it is thus an isomorphism of representations.

  • Let $\newcommand{\Sym}{\operatorname{Sym}} \Sym^2(V, )$ be the linear subspace of $\Bil^2(V, )$ consisting of alternating bilinear forms. This is $$-subrepresentation of $\Bil^2(V, )$. The usual isomorphism of vector spaces $\newcommand{\Symp}{\operatorname{S}} \Sym^2(V, ) ≅ \Symp^2(V)^*$ is an isomorphism of representations. The usual injective linear map $\Symp^2(V^*) \to \Symp^2(V)^*$ is a homomorphism of representations. If $V$ is finite-dimensional, then it is thus an isomorphism of representations.

A bilinear form on a vector space $W$ is the same as a linear map $W \to W^*$. If $W$ is finite-dimensional, then the linear map $W \to W^*$ is an isomorphism if and only if the corresponding bilinear form is non-degenerate. We can do essentially the same for isomorphisms of representations $V ≅ V^*$.

  • Every bilinear form $β$ on $V$ determines a linear map $$ β' \colon V \longrightarrow V^* \,, \quad v \longmapsto β(v, -) \,. $$ The bilinear form $β$ is $$-invariant if and only if this linear map $β'$ is a homomorphism of representations. Consequently, if $V$ is finite-dimensional and $β$ is both $$-invariant and non-degenerate, then $β'$ is an isomorphism of representations $V ≅ V^*$.

As a consequence, we have the following:

Suppose that the representation $V$ is finite-dimensional, and that it admits a non-degenerate, $$-invariant bilinear form $β$.

  1. There exist isomorphisms of representations \begin{gather*} \Bil^2(V, ) ≅ (V ⊗ V)^* ≅ V^* ⊗ V^* ≅ V ⊗ V \,, \\[0.5em] \Alt^2(V) ≅ \Bigl( ⋀^2 V \Bigr)^* ≅ ⋀^2 V^* ≅ ⋀^2 V \,, \\[0.5em] \Sym^2(V) ≅ \Symp^2(V)^* ≅ \Symp^2(V^*) ≅ \Symp^2(V) \,. \end{gather*}

  2. The bilinear form $β$ is an invariant element of $\Bil^2(V, )$, whence $V ⊗ V$ cannot be irreducible unless it the one-dimensional trivial representation. This requires $V$ to be one-dimensional.

  3. Suppose that $β$ is alternating. It it then an invariant element of $\Alt^2(V, )$, whence $⋀^2 V$ cannot be irreducible unless it is one-dimensional and trivial. This requires $V$ to be two-dimensional.

  4. Suppose that $β$ is symmetric. It it then an invariant element of $\Sym^2(V, )$, whence $\Symp^2(V)$ cannot be irreducible unless it is one-dimensional and trivial. This requires $V$ to be one-dimensional.


We can pretty much directly apply this general argumentation to the given situation. We only need to know that the form $Q$ is $_4(ℂ)$-invariant. Depending on the given definition of $_4(ℂ)$, this is either true by definition, or a consequence of the fact that $Q$ is $\mathrm{Sp}_4(ℂ)$-invariant.

  • Thanks for this response. I have one doubt though. This might sound silly, but will you mind explaining how the correspondence occur between the bilinear form being lie algebra invariance and lie group invariance. I mean how do you get one condition from another. – simu tiyam Dec 22 '22 at 16:17
  • Great answer! I had related discussions here: https://math.stackexchange.com/a/3701479/96384. – Torsten Schoeneberg Dec 22 '22 at 17:20
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    @simutiyam That comes from differentiating the Lie group condition: let $g =\exp(tx)$ and differentiate $B(gv,gw)=B(v,w)$ which yields $B(xv,w) + B(v,xw) =0$. Note this is basically just product rule. – Callum Dec 22 '22 at 17:56