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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!

  • Sorry, what does "explicit matrices" mean? You mean like their action in a defining representation? – Ben Webster Apr 19 '22 at 13:50
  • @BenWebster A matrix Lie algebra consists of matrices, no? – theHigherGeometer Apr 19 '22 at 13:54
  • Sure, from one perspective. But there's no one answer, since you have to choose the underlying bilinear form. – Ben Webster Apr 19 '22 at 14:03
  • Take the standard Euclidean inner product on R^n, define the orthogonal group, then take its Lie algebra. Take the usual quaternion-based construction of Sp(n), take its Lie algebra. I'm taking a very pedestrian viewpoint here. No fancy choices of bilinear forms to optimise expressions in the representation theory. – theHigherGeometer Apr 19 '22 at 14:09

2 Answers2

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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.
Ben Webster
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  • Sorry, I should have been specific at the start: I meant over the reals... cf my comment about my background in Lie theory! – theHigherGeometer Apr 19 '22 at 14:02
  • @theHigherGeometer I don't understand what distinction you're making. You just don't like that this uses an indefinite form for $\mathfrak{so}_n$? – Ben Webster Apr 19 '22 at 14:09
  • No, I'd prefer everything coming from compact Lie groups, not reductive or split forms. – theHigherGeometer Apr 19 '22 at 14:13
  • Like I said: "it is less easy to find a reference that gives explicit matrices", as opposed to something over a different field, where one has to then manually change the bilinear form defining the Lie goup, and then use a different basis of the Lie algebra to .... etc. :-) – theHigherGeometer Apr 19 '22 at 15:21
  • I appreciate you continuing to expand your answer, but it seems we have a different notion of "explicit" here, and moreover I gently point out the question is tagged [reference-request]. How hard is it to give an actual formula, rather than things like "looking at a $4\times 4$ block along the diagonal, and picking out one of the factors ...". Perhaps you are setting me an assignment. But if I'm writing for a physics audience, and I want to justify a bunch of normalisations, pointing people towards recipes with implicit rep thy or Lie thy background isn't sufficient. – theHigherGeometer Apr 19 '22 at 23:33
  • @theHigherGeometer The issue, from my perspective, is that what you're asking for doesn't quite make sense, so I can explain what's going on, but I can't give you matrices that don't exist. There are no coroot vectors in the Lie algebra of the compact group, so I can't write formulas for them. – Ben Webster Apr 20 '22 at 14:03
  • Given my ignorance of Lie theory, perhaps I'm using the wrong word. Dual root? – theHigherGeometer Apr 20 '22 at 20:12
  • A combination of the phrases "long root" and "normalise the Killing form" should surely be enough information for an expert to figure out what elements of the Lie algebra I need to check this normalisation :-) Because it is apparently enough in all the literature I've read to convey a precise identification of a unique inner product. Me being inexpert and looking for some specific surely well-known pieces of information so I can identify for myself what this inner product is in some concrete cases. – theHigherGeometer Apr 20 '22 at 21:37
  • As much as I appreciate the extra effort you are putting in, we really are talking past each other. I come from physics background originally, where "explicit" means things like the Pauli matrices or the Gamma matrices. The generality you are supplying is utterly unneeded! And I really mean it when I say I'm looking for a reference to a textbook or a paper. – theHigherGeometer Apr 21 '22 at 00:42
  • @theHigherGeometer what elements of the Lie algebra I need to check this normalisation Yes, like I said above: take any norm-square 2 element of su(2) (one of the Pauli matrices) and look at the norm-square of its image. – Ben Webster Apr 21 '22 at 19:38
  • @theHigherGeometer Obviously, at this point, this in part because I want to understand the answer clearly. For physicists, I would just do the cases of su(2), so(3),so(4) and sp(4), and then say the higher rank cases are the same. – Ben Webster Apr 21 '22 at 19:40
  • Thanks for this, it is eventually sinking in. I think one more question, to clarify: for $\mathfrak{sp}(2)$, i.e. 2 by 2 quaternionic matrices, the long coroot comes from thinking of $\mathfrak{su}(2) \to \mathfrak{sp}(2)$ and taking image of the long coroot? And by long coroot you mean the coroot associated to a long root, right? (Others have called these short coroots, here on MO) – theHigherGeometer Apr 23 '22 at 00:51
  • @theHigherGeometer Yes, the long coroot corresponds to that map su(2) -> sp(2). I say "corresponds" because it's the image of $(\begin{smallmatrix}1 & 0 \ 0 & -1\end{smallmatrix})$ under that map, which makes no sense, because that's not in the domain. But for purposes of computing length, there's so need to use the coroot, you can use any element of su(2), and compare its length in the usual inner product on su(2) to the length of the image. – Ben Webster Apr 23 '22 at 01:10
  • @theHigherGeometer long coroots correspond to short roots: the product of the length of the root and coroot is fixed to be 2 (because that's the answer in sl(2)) so if one is longer, the other is shorter. – Ben Webster Apr 23 '22 at 01:11
  • Sorry, more: "This is the long coroot, so need to normalize so this matches the obvious form on su(2)." I'm not sure what is going on in this sentence. Normalise the coroot? The inner product? What matches the obvious form? And what is this obvious form? – theHigherGeometer Apr 23 '22 at 01:16
  • Hmm. I need the elements of the Lie algebra corresponding to the coroots of the long roots. I'm wary, because of all the possible reciprocals of prefactors that can be involved here. When I'm "comparing lengths", this doesn't tell me what the normalisation factor actually is, only up to inversion. This is why I'm making such a big deal over this, because there's dualisations and inversions and embeddings and pullbacks and so on, and it's never unwound to give an actual number. – theHigherGeometer Apr 23 '22 at 01:28
  • "the image of X under the map su(2)→so(3,R);" <--- except X is an imaginary quaternion, not an element of su(2). Are you using the isomorphism from the first bullet point between Im(H) and su(2) here, or another one? That is, do you mean the image of X_C instead? – theHigherGeometer Apr 23 '22 at 03:40
  • @theHigherGeometer yes, I should have said the image of $X_{\mathbb{C})$ under this map. I corrected in the answer. – Ben Webster Apr 24 '22 at 19:12
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Over on MathOverflow, Konrad Waldorf supplied the reference

and says that section 4 therein

lists, in an absolutely concrete way, the simple Lie algebras realized as matrix Lie algebras, together with all roots and coroots.