# Representations of the symplectic Lie algebra $\mathfrak{sp}_{4}\mathbb C$

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 $⋀^2 V$. Commented Dec 22, 2022 at 11:46
• Yes, you are right. That's a mistake from my end. Commented Dec 22, 2022 at 16:23

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

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

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 $$β \colon V × V \to 𝕜$$ is called $$𝔤$$-invariant if $$β(xv, w) + β(v, xw) = 0 \qquad \text{for all x ∈ 𝔤 and all v, w ∈ V.}$$ If $$𝔤$$ 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 “$$𝔤$$-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 $$𝔤$$ via $$(x β)(v, w) = β(xv, w) + β(v, xw) \qquad \text{for all x ∈ 𝔤, β ∈ \Bil^2(V, 𝕜), v, w ∈ V.}$$ A bilinear form $$β$$ is $$𝔤$$-invariant if and only if it invariant in the general sense.

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

• The usual isomorphism of vector spaces $$\Bil^2(V, 𝕜) ≅ (V ⊗ V)^*$$ 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 $$\newcommand{\Alt}{\operatorname{Alt}} \Alt^2(V, 𝕜)$$ be the linear subspace of $$\Bil^2(V, 𝕜)$$ consisting of alternating bilinear forms. This is $$𝔤$$-subrepresentation of $$\Bil^2(V, 𝕜)$$. The usual isomorphism of vector spaces $$\Alt^2(V, 𝕜) ≅ (⋀^2 V)^*$$ 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 $$\newcommand{\Sym}{\operatorname{Sym}} \Sym^2(V, 𝕜)$$ be the linear subspace of $$\Bil^2(V, 𝕜)$$ consisting of alternating bilinear forms. This is $$𝔤$$-subrepresentation of $$\Bil^2(V, 𝕜)$$. The usual isomorphism of vector spaces $$\newcommand{\Symp}{\operatorname{S}} \Sym^2(V, 𝕜) ≅ \Symp^2(V)^*$$ 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 $$𝔤$$-invariant if and only if this linear map $$β'$$ is a homomorphism of representations. Consequently, if $$V$$ is finite-dimensional and $$β$$ is both $$𝔤$$-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, $$𝔤$$-invariant bilinear form $$β$$.

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

2. The bilinear form $$β$$ is an invariant element of $$\Bil^2(V, 𝕜)$$, 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 $$\Alt^2(V, 𝕜)$$, 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 $$\Sym^2(V, 𝕜)$$, 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 $$𝔰𝔭_4(ℂ)$$-invariant. Depending on the given definition of $$𝔰𝔭_4(ℂ)$$, 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. Commented Dec 22, 2022 at 16:17
• Great answer! I had related discussions here: math.stackexchange.com/a/3701479/96384. Commented Dec 22, 2022 at 17:20
• @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. Commented Dec 22, 2022 at 17:56