The $D_d$ Dynkin diagram otherwise known as SO(2d) one of whose real forms is SO(d,d). |

Let there be d nodes to this Dynkin diagram and let them be numbered along the long leg from left to right 1 to (d-2), and for the two fish-tail nodes let the bottom one be number (d-1) and the top one node d.

One can form a spinorial representation of SO(d,d) by attaching an extra node, which we will number (d+1), to node (d-1) on the diagram above and considering all the roots associated to the extended Dynkain diagram such that the root $\alpha_{(d+1)}$ appears only once. This has the effect of constructing the representation of SO(d,d) with lowest weight $-\lambda_{(d-1)}$. Usually we work with highest weight representations, in this construction we work from the bottom up building on the lowest weight. This representation will be the spinorial representation.

So far not so much fun... But we may well wonder how big is this representation? To this end let us decompose the extended Dynkin diagram to tensor representations of SL(d,${\mathbb R}$) by deleting two nodes. Recall that in order to build the spinorial representation we added node (d+1), which is not shown and held it fixed - there was always one multiple $\alpha_{(d+1)}$ in any root of this representation - well now we will delete this and we will also delete node (d). Deleting node (d+1) gives the vector representation of SL(d, ${\mathbb R}$) of dimension $d$ while deleting node (d) gives the antisymmetric 2-index SL(d, ${\mathbb R}$) tensor of dimension $\frac{d(d-1)}{2}$. We can usefully denote these two representations by Young tableaux:

Generically the indices are denoted $a$ and $b$ but can range from 1 to (d). In fact upon deletion of the nodes these tableaux takes specific values $a=d$ and $b=(d-1)$. The dimensions of the representations mentioned above are clear when we let the values $(a,b)$ take all possible values allowed by the symmetry of the tableaux.

Now as we construct the spinorial representation more and more Young tableau will appear. How can we tell which ones will show up? As SO(d,d) has a finite-dimensional Lie algebra and as the Dynkin diagram is simply-laced, all roots have the same length and for a given (d) there are a finite number of them. Let each root have root length squared equal to 2. We can embed the roots into a vector space $V_{(d+1)}$ with basis elements $e_1,e_2,e_3\ldots e_{(d-1)}, e_d, e_{(d+1)}$. To do this we must preserve all the inner products encoded in the Dynkin diagram between the simple positive roots - this amounts to us being able to find an inner product which will achieve this. Let the simple positive roots in $V_{(d+1)}$ be

$$\alpha_{i}=e_i - e_{i+1} \qquad \qquad (1\leq i \leq (d-1))$$

$$\alpha_d = e_{(d-1)}+e_d+e_{(d+1)}$$

$$\alpha_{(d+1)}=e_d-e_{(d+1)}$$

Under the usual scalar product $\alpha_d^2=3$ while all the other roots have squared length 2, as desired. We therefore modify the inner product to be given by:

$$<\beta,\gamma>=\sum_{i=1}^{d+1}b_ic_i-(m_d)_\beta(m_d)_\gamma$$

Where $\beta=\sum_{i=1}^{d+1}b_ie_i=\sum_{i=1}{d+1}m_i\alpha_i$ and $\gamma=\sum_{i=1}^{d+1}c_ie_i$ and $(m_d)_\beta$ is the number of times the root $\alpha_d$ appears in the simple root expansion of $\beta$. So now all roots have length squared 2 and the inner products are those corresponding to the simple roots of our extended SO(d,d) diagram. For reference one can work out the fundamental weights of SO(d,d) in this vector space basis and for this inner product it is the vector with components $-\frac{1}{2}(1,1,\ldots 1,1,-1,(2-d))$ i.e. it is:

$$\lambda_{d+1}=-\frac{1}{2}(e_1+e_2+\ldots + e_{(d-2)}+e_{(d-1)})+\frac{1}{2}e_{(d)}+\frac{(d-2)}{2}e_{(d+1)}$$

This looks rather odd but then we are making a peculiar spinor construction by embedding the SO(2d) root lattice inside that of $E_{(d+1)}$. However you can see in the first $d$ entries the usual highest weight for the spinorial representation, see the more usual discussion (without the embedding in $E_d$) in Wikipedia for example (look at the section spin representations and their weights). The more familiar story and the connection to Clifford algebras follows from here. But we continue down our path less travelled...

$$\lambda_{d+1}=-\frac{1}{2}(e_1+e_2+\ldots + e_{(d-2)}+e_{(d-1)})+\frac{1}{2}e_{(d)}+\frac{(d-2)}{2}e_{(d+1)}$$

This looks rather odd but then we are making a peculiar spinor construction by embedding the SO(2d) root lattice inside that of $E_{(d+1)}$. However you can see in the first $d$ entries the usual highest weight for the spinorial representation, see the more usual discussion (without the embedding in $E_d$) in Wikipedia for example (look at the section spin representations and their weights). The more familiar story and the connection to Clifford algebras follows from here. But we continue down our path less travelled...

We were building up the spinorial representation for which we held $m_d=1$ for all the roots in the representation. We may classify the roots that appear by the number of copies $\alpha_d$ they possess, and that number itself we call the level. So at level $m_d=0$ we find just

having dimension d. At level 1 we consider the tensor product and decompose it using the Littlewood-Richardson rules to find:

There are two possible Young tableau at level one but in fact only the first has root length squared equal to two, the second has length squared equal to four. As the roots in the representation all have length squared two only the first Young tableau can exist in the algebra. In fact as we keep going and construct the possible Young tableaux at level two, three, four... only the completely antisymmetric tableaux consisting of $2m_d+1$ boxes have length squared two - as a quick computation using the inner product shows.

Finally we have a closed statement that the spinorial representation of SO(d,d) can be represented by a sum of SL(d,$\mathbb R$) antisymmetric tensors of $2m_d+1$ indices. For a finite d this sum of tensors terminates when either $2m_d+1=d$ for odd d, or when $2m_d+1=d-1$ for even d:

For the even d case the dimensions of the $m_d+1=\frac{d}{2}$ Young tableaux may be summed to

For the even d case the dimensions of the $m_d+1=\frac{d}{2}$ Young tableaux may be summed to

$$d+\frac{d(d-1)(d-2)}{3!}+\frac{d(d-1)(d-2)(d-3)(d-4)}{5!}+...+d=\sum_{i=0}^{\frac{d}{2}-1}{ d\choose (2i+1)}=2^{(d-1)}$$

While for odd d there are $\frac{d+1}{2}$ tableaux giving:

$$d+{d\choose 3} + \ldots + {d\choose (d-2)}+{d\choose d}=2^{(d-1)}$$

N.B. for the sums one can use the neat trick of adding or subtracting $0=(1-1)^d$ to $2^d=(1+1)^d$ to prove the two sums above are the same. (Thanks to V. for pointing this out.)

Et voila! In both even and odd d we count $2^{(d-1)}$, if we happened to not be interested in SO(d,d) but rather SO(D), where D=2d is even we find the dimension of the spinorial representation is $2^{\frac{D}{2}-1}$. This is one of the two inequivalent Weyl spinor representations (there is one associated to each node in the fish tail of the Dynkin diagram) and together they give a Dirac spinor of dimension $2^{\frac{D}{2}}$. For odd D one has to start with the $B_n$ Dynkin diagram and that's another story...

N.B. for the sums one can use the neat trick of adding or subtracting $0=(1-1)^d$ to $2^d=(1+1)^d$ to prove the two sums above are the same. (Thanks to V. for pointing this out.)

Et voila! In both even and odd d we count $2^{(d-1)}$, if we happened to not be interested in SO(d,d) but rather SO(D), where D=2d is even we find the dimension of the spinorial representation is $2^{\frac{D}{2}-1}$. This is one of the two inequivalent Weyl spinor representations (there is one associated to each node in the fish tail of the Dynkin diagram) and together they give a Dirac spinor of dimension $2^{\frac{D}{2}}$. For odd D one has to start with the $B_n$ Dynkin diagram and that's another story...