# What is Pressley and Segal's "basic inner product" for compact simple Lie algebras of types B and C?

In Pressley and Segal's book Loop Groups, they define a "basic inner product" $$\langle-,-\rangle$$ on a simple Lie algebra to be (minus) the Killing form scaled so that $$\langle h_\alpha,h_\alpha\rangle=2$$ where $$h_\alpha$$ is the coroot associated to a long root.

[Aside I believe there is some terminological confusion possible here: they surely mean $$h_\alpha$$ to be an element of the Lie algebra, rather than an element of $$\mathfrak{t}^*$$; Terry Tao defines a coroot to be such a thing, appropriately scaled. But others take a coroot to be a rescaled root, for instance Wikipedia.]

They give the examples of $$\mathfrak{su}(n)$$ and $$\mathfrak{so}(2n)$$, which are $$-\mathrm{tr}(XY)$$ and $$-\frac12\mathrm{tr}(XY)$$, respectively. To avoid confusion, this trace is taken in the defining representation, thinking of these as matrix Lie groups.

But I haven't been able to find a source that gives the basic inner product for $$\mathfrak{so}(2n+1)$$ or $$\mathfrak{sp}(n)$$, where the latter are given as $$n\times n$$ quaternionic matrices. Once one knows the appropriate matrices $$h_\alpha$$, then it is obvious. And it is sufficient to know these for low ranks, for instance $$\mathfrak{so}(5)$$, since then matters stablise.

I tried asking for these actual matrices over at M.SE, but despite rather a detailed answer, I still have no joy, as it presumes a lot of "common knowledge" of Lie theory I don't have, and also is slightly loose with terminology and suggestions. I was hoping for something super explicit like writing out the analogues of the Pauli matrices. This seemed like material too basic for MO, but despite a lot of searching, no lecture notes or textbook I've found actually gives this information! Nor have I found student exercises that ask for them. To me this seems like a perfect example of what MO was intended for: a place for researchers to ask colleagues about a basic fact they need from an area they are unfamiliar with. So I'm cutting my losses, forgetting the exercise of trying to work this stuff out for myself, and asking outright:

What is the basic inner product on $$\mathfrak{so}(2n+1)$$ and $$\mathfrak{sp}(n)$$?

My motivation is that eventually I'm going to be doing physics-style calculations, and need explicit representatives for absolutely everything. So a characterisation in terms of anything else is insufficient: I want a formula. A reference to a place where this is recorded in the literature would be the best answer, but I despair that such a thing exists.

• Isn't this effectively just asking for the normalization of the quadratic Casimir operators for those algebras? All the necessary quantities are probably to be found here: scipp.ucsc.edu/~haber/ph251/Casimir2.pdf
– Buzz
Commented Apr 24, 2022 at 3:55
• @Buzz apologies if that sounds harsh, I shouldn't be taking out my frustrations, generated elsewhere, on users here! Commented Apr 25, 2022 at 12:48
• No problem; I see much more bitter complaints on the Physics Stack Exchange site all the time.
– Buzz
Commented Apr 25, 2022 at 14:43
• @ulnor Tao explicitly writes "$H_\alpha$ is the co-root of $\alpha$, defined as the element of $\mathfrak h$ given by the formula ...", but is taking roots to live in $\mathfrak{h}^*$. Check for yourself :-) And given that WP defines roots to live in $\mathfrak{h}^*$, and coroots to be scaled roots in the same vector space (as at the link I supplied), then coroots live in $\mathfrak{h}^*$, as you point out. The source Konrad supplies defines coroots to live in the original Lie algebra, but many other sources define coroots to be in the dual of the Cartan subalgebra. So... :-) Commented Apr 26, 2022 at 7:51
• @ulnor I checked, and Tao only gives a formula saying coroots are in the same space as roots after using the Killing form to construct an isomorphism $\mathfrak{h}^*\simeq \mathfrak{h}$ (this simplifies the formula he gives that defines the coroots (written "co-roots" on the page; he uses both forms). So there is legitimately some kind of convention in both directions in the literature, I would guess. Commented Apr 26, 2022 at 7:58

## 2 Answers

Section 4 of

Gawȩdzki, Krzysztof; Reis, Nuno, Basic gerbe over non-simply connected compact groups, J. Geom. Phys. 50, No. 1-4, 28-55 (2004). ZBL1067.22009.

lists, in an absolutely concrete way, the simple Lie algebras realized as matrix Lie algebras, together with all roots and coroots. From these you can get the normalization.

For example, let's look at $$\mathfrak{so}(2n+1)$$. Let $$e_i$$ be the $$(2r+1)\times (2r+1)$$ matrix with a block $$\begin{pmatrix}0 &-1\\1& 0\end{pmatrix}$$ on the diagonal at the $$(2i-1)$$th and $$2i$$th position. Then, the roots are the matrices $$\pm e_i \pm e_j$$ for $$i\neq j$$ and $$\pm e_i$$, and the coroots are $$\pm e_i \pm e_j$$ for $$i\neq j$$ and $$\pm 2e_i$$. You can check that $$\mathrm{tr}(X^2)=-4$$ for all coroots $$X$$. Thus, the basic inner product for $$\mathfrak{so}(2n+1)$$ is $$-\frac{1}{2}\mathrm{tr}(XY)$$ just as in the even case.

• Oh, excellent, thanks! Commented Apr 25, 2022 at 12:35
• The remaining problem is thus to get the type C one, using quaternionic matrices. Section 4.3 has "With the invariant form normalized so that..." and then gives the roots relative to that. I guess working out the case of sp(2) shouldn't be too awful... If you feel like also giving this as an answer at math.stackexchange.com/questions/4430933/…, that would be awesome. Commented Apr 25, 2022 at 12:46
• My only worry is this: "We identify in the standard way t and g with their duals using the ad-invariant bilinear form tr XY on g. The normalization of tr is chosen so that the long roots viewed as elements of t have length squared 2", so that the isomorphism between t and t* might not be not the usual one that doesn't build this in. Perhaps my worries are unfounded...? Commented Apr 25, 2022 at 13:29
• Here's something that confuses me: why are the long and short (dual) roots the same length in this metric on so(2n+1)? No, wait, how can you have tr(XY)=-4 always? Some of them should be orthogonal, right? What are your X and Y in this equation? Seems like a typo. Commented Apr 25, 2022 at 21:56
• @DavidRoberts: yes, there was a typo, I meant $\mathrm{tr}(X^2)=-4$. Commented Apr 26, 2022 at 5:50

I found a reference that gave the correct inner products explicitly, without requiring the reader to assemble the relevant facts: Chapter II, section 1.2 (bottom of page 583) of:

• McKenzie Y. Wang, Wolfgang Ziller, On normal homogeneous Einstein manifolds, Annales scientifiques de l'École Normale Supérieure, Série 4, Tome 18 (1985) no. 4, pp. 563-633. https://doi.org/10.24033/asens.1497

Giving:

• $$\mathfrak{su}(n)$$: $$-\mathrm{tr}(AB)$$
• $$\mathfrak{so}(n)$$ ($$n\geq 5$$): $$-\frac12 \mathrm{tr}(AB)$$; and for $$\mathfrak{so}(3)$$: $$-\frac14 \mathrm{tr}(AB)$$
• $$\mathfrak{sp}(n)$$: $$- \mathrm{Tr}(AB)$$

Here the trace $$\mathrm{Tr}$$ for quaternion matrices is the reduced trace, namely $$\mathrm{Tr}(X) = 2\Re \mathrm{tr}_{\mathbb{H}}(X)$$, which can be checked by looking at the Dynkin index of the embedding $$\mathfrak{sp}(2) \hookrightarrow \mathfrak{su}(4)$$, which is 1, and the index of the standard embeddings $$\mathfrak{sp}(n) \hookrightarrow \mathfrak{su}(n+1)$$ (also 1).