# Why Lie algebras of type $B_2$ and $C_2$ are isomorphic?

both of Lie algebras of type $B_2$ and $C_2$ have dimension 10 and we can find two basis of them on page 3 in the book: Introduction to Lie algebras and representation theory . How could we show that these two Lie algebras are isomorphic by constructing an explicit correspondence between two basis of them. Thank you very much.

## 1 Answer

$B_2=so(5)$ and $C_2=sp(4)$. Let me write it as $C_2=sp(V)$ where $V$ is a 4-dim. symplectic vector space with sympl. form $\omega$. The 6-dim. vect. space $\wedge^2 V^*$ has a natural inner product given by the wedge product: $(\alpha,\beta)$ is defined by $\alpha\wedge\beta=(\alpha,\beta)\,\omega\wedge\omega$. Let $W\subset \wedge^2 V^*$ be the orthogonal complement of $\omega\in\wedge^2 V^*$. $W$ is 5-dimensional and has an inner product. The action of $sp(V)$ on $V$ yields an action on $W$ preserving the inner product, i.e. to a Lie algebra morphism $sp(V)\to so(W)$. And this morphism is in fact an isomorphism.

• I am sorry that I could not understand some parts of the answer. Here $\omega$ is a linear map from $V \times V$ to $\mathbb{C}$. It is said that $\omega \in \wedge^2 V^*$. Is this because we could identify $V$ with $V^*$? What is the action of $sp(V)$ on $W$ explicitly? Why $sp(V) \to so(W)$ is isomorphic? – user Mar 26 '11 at 15:40
• $\wedge^2 V^*$ is by definition (well, if $V$ is finite-dimensional) the space of skew-symm. bilin. forms on $V$. If $A\in sp(V)$ and $a\wedge b\in\wedge^2 V^*$ then $A\cdot (a\wedge b)=-(aA)\wedge b - a\wedge(b A)$ (here $a\mapsto -aA$ is the action on $V^*$). The morphism $\mu:sp(V)\to so(W)$ is not 0 (has to be checked for one $A\in sp(V)$). Since $sp(V)$ is simple, the kernel is either everything or $0$, so it's $0$ and $\mu$ is injective. It's a bijection since $sp(V)$ and $so(W)$ have the same dimension (not very elegant, but I don't know a better explanation). – user8268 Mar 26 '11 at 22:40
• @user8268, thank you very much. Why the action of $sp(V)$ is defined in this way but do not take other actions? Here $a\in V^*$ is a linear function from $V$ to $\mathbb{C}$ and $A \in sp(V)$ is a matrix. What is the result of $aA$? I think $aA$ should be a linear function from $V$ to $\mathbb{C}$. What is this linear function? Why the morphism $\mu: sp(V) \to so(W)$ is not 0? Do we need the property that $W$ is the orthogonal component of $\omega$ to show this? – user Mar 26 '11 at 23:16
• $(aA)(v)=a(A(v))$. To show $\mu\neq0$ choose the simplest non-0 $A\in sp(V)$ and show that $\mu(A)\neq0$. – user8268 Mar 27 '11 at 6:47
• But where do we need the condition that $W$ is the orthogonal complement of $\omega$? – user Mar 28 '11 at 0:28