Explicit matrices $h_\alpha$ that correspond to the long roots $\alpha$ in a classical compact simple Lie algebra over the reals

Question

While it is easy to find many sources that give expressions for the (co)roots of an abstract root system, it is less easy to find a reference that gives explicit matrices that are the "coroots" (in the sense of Lie algebra elements) in the simple classical Lie algebras (i.e. types $A_n$, $B_n$, $C_n$ and $D_n$). I'm particularly interested in those corresponding to the long roots, but a reference that gave all of them would be perfect. Edit: to clarify, I mean the compact forms of the real matrix Lie algebras.

The reason I ask is that it is useful to normalise the Killing form so that the these algebra elements have length $\sqrt{2}$, and I want to give these normalisations as examples for the classical Lie algebras for a paper I'm writing, since sources I'm familiar with just specify the case $\mathfrak{su}(n)$. For the non-simply-laced cases it is less easy to sort out what's going on, as my background in Lie theory is weak, so chasing the definitions from the abstract root system through the corresponding decomposition of the Lie algebra etc is not obvious. But a source that just gives the answer is not forthcoming after a decent internet search!

Answer 1

This is in Fulton and Harris. Look on page 240-1 for $\mathfrak{sp}$ and page 270-1 for $\mathfrak{so}$.

EDIT: Apparently the issue here is that the OP wants to change basis to the standard version of the bilinear forms. For $\mathfrak{sp}_{2n}$, there's no issue here. All real symplectic forms are equivalent. If you want to think about the compact form as transformations of $\mathbb{H}^n$ preserving quaternionic norm, Fulton and Harris are thinking about this as a complex vector space with basis $e_1,\dots, e_n, je_1,\dots, je_n$.

More explicitly, the Lie algebra is the anti-Hermitian quaternionic $n\times n$ matrices. The "obvious" Cartan is given by diagonal matrices with values in $i\mathbb{R}$ (note: nothing special about $i$; conjugating by unit quaternions sends this to real multiples of any imaginary quaternion), and the SU(2)'s for the simple roots are given by:

• the block diagonal $2\times 2$ complex anti-Hermitian matrices in a consecutive pair of rows and columns. The long coroots are the images of $\begin{bmatrix}1 & 0\\ 0& -1\end{bmatrix}$ along this diagonal. Under the usual trace form on quaternion matrices, this has inner product 2 with itself (so inner product 4 if you use the action on $\mathbb{C}^{2n}$ instead).
• the diagonal matrices $\mathrm{diag}(0,\dots,0,u)$ for $u$ an imaginary quaternion. The inner product on this under quaternionic trace form is half the usual inner product on $\mathfrak{su}(2)$ (which gives Pauli spin matrices norm $\sqrt{2}$), so the corresponding coroot has inner product 1 with itself.

For $\mathfrak{so}_{n}$, you need to change the real form to pair together the basis vectors that pair non-trivially. The torus of $\mathfrak{so}_{n}$ acts on these real 2-d spaces by the usual rotation. That is, they are block diagonal with $2\times 2$-blocks given by $$\begin{bmatrix}\cos \theta_i & \sin \theta_i\\ -\sin \theta_i & \cos \theta_i \end{bmatrix}$$ The root SU(2)'s come from looking at a $4\times 4$ block along the diagonal, and picking out one of the factors of $SU(2) x SU(2) =Spin(4)$ with the exception of:

• if $n$ is even, in the last block you also take the other SU(2) factor in $Spin(4)$. The usual trace form on all of these is the trace form of imaginary quaternions acting on $\mathbb{H}=\mathbb{R}^4$ as a real vector space, i.e. twice the usual form on $\mathfrak{su}(2)$, so inner product 4.
• if $n$ is odd, then you use the map $SU(2) \to SO(3)$ to act on the last $2\times 2$ block and the odd man out column/row. This is the long coroot, since the coroot in $\mathfrak{so}(3)$ is the matrix $\mathrm{diag}(2,0,-2)$. Note that the inner product of this with itself is 8.

EDIT AGAIN:. I like a good challenge, though this is all getting a little complicated (you have read the twitter threads too if you want all the details). Coroot vectors in the Lie algebra of the compact group don't make sense, so I think it's better to think about homomorphisms of $\mathfrak{su}(2)$ into your Lie algebra. You can think of any unit imaginary quaternion in $\mathfrak{su}(2)$ (for example, any of the Pauli spin matrices) as a coroot vector if you want; the norm-square of the coroot under your form is the pairing of this vector with itself (up to sign). As I said on twitter, there are just a few basic building blocks of these, so let me write those out. Let $X$ be an imaginary quaternion:

• Let $X_{\mathbb C}$ be the $2\times 2$ matrix over $\mathbb C$ given by left multiplication on $\mathbb{H}\cong \mathbb{C}^2$ (i.e. the usual isomorphism to $\mathfrak{su}(2)$). This sends $i.j,k$ to the Pauli matrices. For example, $i$ goes to $(\begin{smallmatrix} i & 0\\ 0 & -i \end{smallmatrix})$.
• Let $X_{\mathbb R}$ be the $4\times 4$ matrices over $\mathbb R$ given by left multiplication on $\mathbb{H}\cong \mathbb{R}^4$.
• Finally, let $X_{\times }$ denote the image of $X$ under the usual map of imaginary quaternions to $\mathfrak{so}(3,\mathbb R)$ (you can also think of this as the map $\mathfrak{su}(2)\to \mathfrak{so}(3,\mathbb R)$); you can think of this as the matrix of taking cross product with $X$ (or equivalently, the matrix given by taking commutator of imaginary quaternions with $X$).

It is an assignment to you to write these in your preferred notation.

• In $\mathfrak{su}(n)$, all root $\mathfrak{su}(2)$'s come from taking $X_{\mathbb C}$ as a diagonal $2\times 2$ block and surrounding it by $0$'s.
• In $\mathfrak{sp}(2n)$, written as quaternionic matrices, all root $\mathfrak{su}(2)$'s come from taking $X$ as a diagonal entry and surrounding it by $0$'s, or taking $X_{\mathbb C}$ as a diagonal $2\times 2$ block and surrounding it by $0$'s (thinking of $\mathbb C$ as a subalgebra of the quaternions); the latter is the long coroot.
• In $\mathfrak{so}(n)$, some root $\mathfrak{su}(2)$'s come from taking $X_{\mathbb {R}}$ as a diagonal $4\times 4$ block and surrounding it by $0$'s (in terms of $\mathfrak{su}(2)$, this means thinking of $\mathbb{C}^2$ as $\mathbb{R}^4$).
• In $\mathfrak{so}(2n+1)$, you also need $X_{\times}$ as a diagonal $3\times 3$ block. This is the long coroot, so need to normalize so this matches the obvious form on $\mathfrak{su}(2)$.
• In $\mathfrak{so}(2n)$, you also need the $4\times 4$ matrix given by right multiplication by $X$ on the quaternions as a diagonal block.