I'm going through some lecture notes I found online about Symmetry for Physicists and I got stuck in a problem of representations and tensors. The problem reads "The Lie algebra so(11) of rotations in 11 dimensions has an irreducible 55-dimensional representation.

Describe the representation module in terms of tensors." I've been thinking about it for a while but I'm not able to relate it to the simpler cases I found in the notes. Could someone provide me with a starting point?

Thank you

# Representation of so(11) in terms of tensors

group-theorylie-algebrasrepresentation-theory

#### Related Solutions

$\newcommand \G{\mathbf{G}}$ $\newcommand \GL{\mathrm{GL}}$ $\newcommand \SL{\mathrm{SL}}$ $\newcommand \Z{\mathbb{Z}}$ $\newcommand \Q{\mathbb{Q}}$ Just to get this out of the unanswered queue, let me give the following answer:

1) Yes! There are actually tons. Any book on 'linear algebraic groups' will cover what you want (although Waterhouse's book is strange it sidesteps a lot of the theory). Specifically though, I would recommend these notes of Milne--I think they are about as good as one could possibly hope for in terms of completeness. They have the downside of being somewhat cagey about using modern algebraic geometry, but only in language. Namely, Milne does talk about nilpotents (which are pivotal in the characteristic $p$ theory since such simple maps like $\text{SL}_p\to\text{PGL}_p$ have non-reduced kernel) but he insists of talking about $\text{MaxSpec}(A)$ instead of $\text{Spec}(A)$ for some unbeknownst reason--it doesn't make a difference, but it's worth noting.

I am also always eager to rep Brian Conrad's notes (for most things). Specifically, there are these from a first course on linear algebraic groups. They have the strong plusses of being written from a 'modern algebro-geometric viewpoint' (e.g. quotients are defined as quotient fppf sheaves, which is what they should be defined as), but falls prey to the neuroticism that befalls all live-TeXed notes--they're fairly all over the place. Unfortunately, that doesn't really cover what you're asking for because it doesn't cover the structure theory of reductive groups (and their representations). For that you'll have to look at notes from his follow-up course. Again, these are probably my favorite notes for the topic, but are somewhat hard to quickly look things up in (something Milne's notes excel in).

Finally, if you want a 'quick fix' you can look at the first (~50 page) section of these notes of Brian Conrad--I think the same material is roughly contained in the appendix to his book (with Prasad and Gabber) *Pseudo-reductive Groups*.

Let me just note that if you're really only interested in the groups you've mentioned (over $\mathbb{Q}$), then everything is easily describable.

For $\mathbf{G}_m$ (everything in here will be over $\mathbb{Q}$) representations $\mathbf{G}_m\to\GL(V)$ correspond to $\Z$-gradings $\displaystyle V=\bigoplus_{i\in\Z}V_i$ where $\G_m$ acts on $V_i$ by the character $z\mapsto z^i$. In particular, every representation is semisimple and the simple representations are just characters (which are precisely the maps $z\mapsto z^i$).

For $\SL_2$ things are ever so slightly more complicated. Namely, $\SL_2$ is reductive (in fact, semisimple) and thus every algebraic representation of $\SL_2$ is semisimple. Thus, we really only need to describe the simple representations of $\SL_2$. To do this, note that $\SL_2$ naturally acts on homogenous polynomials of degree $m$ in the variables $x,y$ (by $(a_{ij})f(x,y):=f(a_{11}x+a_{12}y,a_{21}x+a_{22}y)$, call this representation $V_m$. Then, every simple representation of $\SL_2$ is isomorphic to precisely one of these $V_m$.

(Note that it's not a coincidence that $V_m\cong \mathcal{O}(m)(\mathbb{P}^1)$ since $\mathbb{P}^1$ is the flag variety $\SL_2/B_2$ associated to $\SL_2$ and the Borel-Weil theorem describes a precise relation between representations of $\SL_2$ and (certain) line bundles on the flag variety--this works more genenerally for a semisimple group).

Finally, $\GL_2$ is also reductive (but not semisimple), thus to describe its representation theory we need only describe its simple representations. These come in two flavors. Namely, there is the irreducible tautological representation of $\GL_n$ (acting on $k^n$ in the usual way) and there are the representations $\wedge^i(k^n)$ for $i\leqslant n$. That's all of them.

Of course, to see where these came from--the real answer is the theory of dominant weights. That said, it's always easier to think about the analogy you know from the representation theory of a finite group. Namely, if $G$ is a finite group and $\text{Reg}(G)$ denotes the left regular representation of $G$, in other words $G$ acting on the group algebra $\mathbb{C}[G]$, then there is a decomposition

$$\text{Reg}(G)\cong\bigoplus_{\rho\in\text{Irr}(G)}\rho^{\dim\rho}$$

where $\text{Irr}(G)$ is the set of isomorphism classes of irreducible algebraic representations of $G$. A similar thing happens for a reductive group $G$. For example, for $\GL_n$ one has that the coordinate ring of $\GL_n$ (as a variety) decomposes as a $\GL_n\times\GL_n$-rep (via the multiplication map) as a direct sum of $V\boxtimes V^\ast$ where $V$ runs over the irreducible representations of $\GL_n$. So, see if you can use this to sort out what happens at least for $\G_m$.

2) As I mentioned in the comment, this really depends on what $\Q$-algebras you pick. For example, you know from the Yoneda philosophy that any representation $\rho:G\to\GL_n$ of an algebraic group $G$ is determined by the group maps $G(R)\to\GL_n(R)$ as $R$ varies over all $\Q$-algebras. In, fact it suffices to think about it for $R=\mathcal{O}_G(G)$ which gives you the notion of a comodule.

Of course, things are going to be bad if you want to consider just the values of the representation on some random $R$, even for examples $R=\Q$ I think. I don't have an example off-hand to be honest (there's probably not a hard one) but if you look anything other than $\Q$, say $\Q(i)$, you'll get lots of representations of $\GL_2(\Q(i))$ (say) that won't be algebraic--think about the one that is induced from the non-trivial automorphism of $\Q(i)$.

Lets remark on what it means for a representation of the Virasoro algebra to be unitary. There are mathematical propositions involved in this, but at the core its a definition coming from phyiscs.

The Virasoro algebra enters like this: You have a symmetry group being given by the group of conformal maps on a two-dimensional signature $(1,1)$ space-time like $S^{1,1}$ or $\Bbb R^{1,1}$. This group of symmetries then necessarily has an action on the physical space of states. This means that there is a projective representation on $\Bbb P(H)$, the projective space of some Hilbert space.

Now its a lot easier to deal with unitary representations onto $H$ than projective representations onto $\Bbb P(H)$, thankfully there is a mathematical theorem telling you that for any group $G$ you can lift a projective representation on $\Bbb P(H)$ to a unitary representation of a central extension of $G$ by $U(1)$ on $H$. Unitary means that the elements of the group act by unitary maps.

In our case the situation is like this:

- The group of conformal symmetries is a big infinite dimensional thing. You have $\mathrm{Conf}^0(S^{1,1})\cong \mathrm{Diffeo}_+(S^1)\times \mathrm{Diffeo}_+(S^1)$.
- The Lie algebra of $\mathrm{Diffeo}_+(S^1)$ is the space of vector fields on $S^1$, here given by the closure of the real span of the fields $\frac d{d\theta}, \cos(n\theta)\frac d{d\theta}, \sin(n\theta)\frac d{d\theta}$, $n\in\Bbb N$.
- Note that the complexification of the span of those vector fields is the Witt algebra ($L_n = (-i\cos(n\theta)-\sin(n\theta))\frac d{d\theta}$ and $L_{-n}=(-i\cos(n\theta)+\sin(n\theta)) \frac{d}{d\theta}$ ).
- The Lie algebra of a central extension of the group is a central extension of the original Lie algebra.
- Hence if you have a projective representation of $\mathrm{Conf}^0(S^{1,1})$ you have a representation of (two copies of) the above Lie algebra so that its elements exponentiate to something unitary, i.e. are anti-selfadjoint.
- Since the Hilbert space is
*complex*, you may extend the Lie algebra and the representation to their complexifications.

Now one more ingredient enters: The complexification process and the process of central extension commute for Lie algebras. Hence we could complexify before doing the central extension and thus what we get is a complex representation of the central extension of the Witt algebra, which is also known as the Virasoro algebra.

Now we need to understand how the condition of unitary enters. We understood what it means in point 6. above. We had some Lie algebra $\mathrm{Vir}_R$ (whose complexification $\mathrm{Vir}_R^{\Bbb C}$ is the Virasoro algebra) so that $\mathrm{Vir}_R$ is mapped by the representation to anti-selfadjoint (or at least anti-symmetric) operators. Concretely we know that $L_n = -i\cos(n\theta) \frac{d}{d\theta} - \sin(n\theta)\frac{d}{d\theta}$, hence $$\rho(L_n)^* = \rho(-i\cos(n\theta)\frac d{d\theta})^* -\rho(\sin(n\theta)\frac{d}{d\theta})^*= (\overline{-i}) \rho(\cos(n\theta)\frac d{d\theta})^* - \rho(\sin(n\theta)\frac d{d\theta})^* \\ = (i)(-\rho(\cos(n\theta)\frac d{d\theta}) + \rho(\sin(n\theta)\frac d{d\theta})= \rho(L_{-n}).$$ Additionally going through the process you should find that the central charge must be symmetric in the representation.

To summarise:

There is a real form $\mathrm{Vir}_R\subset \mathrm{Vir}$ so that $\rho(X)^*=-\rho(X)$ for all $X\in \mathrm{Vir}_R$, that is the Lie-algebra representation on $\mathrm{Vir}_R$ is unitary in the usual sense. Since $\mathrm{Vir}$ is the complexification of this algebra, the extension of the representation to $\mathrm{Vir}$ is not unitary in that sense, but still unitary in the sense that it is coming from a unitary representation of some group.

There are more interesting things to be said. For example while (the completion of) $\mathrm{Vir}_R$ is real Lie algebra of some real infinite dimensional group, there doesn't exist any complex group having $\mathrm{Vir}$ as its complex Lie algebra. Another interesting point I like a quite a bit is remark 3 on page 4 of these elementary notes on the subject.

## Best Answer

The defining representation generators of

SO(n)aren×nantisymmetric real matrices $M_{ij}$, so, then,n(n-1)/2of them, wherei,j=1,...,n. Recall the generators you know fromn=3.For

n=11, you then have 55 such matrices, the dimension of theso(11)algebra. The real structure constants of the commutators of these 55 matrices, $M^a_{ij}$, now indexed bya=1,....,55, among themselves$^1$, are the representation matrices of the adjoint representation, $f^a_{bc}$. They act on the space of 55-vectors, easy to construct from the rank-2 antisymmetric tensors $M_{ij} $.^{1}_{ $[M_b,M_c]=f^a_{bc}M_a.$ }