I am working my way through "Basic Concepts of String Theory", by Blumenhagen, Lüst and Theisen. Currently I am working on the compactifications of string theories on Calabi-Yau manifolds.

Part of the motivational discussion for such compactifications is about the amount of supersymmetry after dimensional reduction. Two statements:

"One way to obtain four-dimensional supersymmetric theories is to start in $d_c =10$ or $d_c = 11$ (here $d_c$ is the critical dimension, or total dimension) and compactify on a $D$-dimensional torus, keeping only the constant modes on the torus, i.e. those which do not depend on the torus coordinates $y_m$. This is referred to as dimensional reduction. The reason why we consider also $d_c=11$ is the existence of a unique supergravity theory in this dimension. [...] A Majorana-Spinor $\psi$ in $d_c=11$ decomposes under $SO(3,1)\times SO(7)$ as $\mathbf{32 = (4,8)}$, where $\mathbf{4}$ and $\mathbf{8}$ are Majorana spinors of $SO(3,1)$ and $SO(7)$, respectively. Hence, $\psi$ gives rise to eight Majorana spinors in $d=4$. This means that starting with $d_c=11$, $\mathcal{N} =1$, in which the supercharge $Q$ is Majorana, gives a $d=4$, $\mathcal{N}=8$ theory upon dimensional reduction".

I have had very limited prior exposure to supergravity theories, and it is not the main topic I have to study right now. However, this means that I do not quite understand the above statement. Some questions:

  • How can I see that a 11-dimensional Majorana spinor decomposes under $SO(1,3) \times SO(7)$? (and what does $\mathbf{32 = (4,8)}$ mean? I have a vague intuitive idea only).

  • Why do we get $\mathcal{N} = 8$ supersymmetry in 4 dimensions? I mean, somehow this needs to be extracted from the $\mathbf{4}$ above (which is the '4d'-part of the decomposition), but I do not see how.

  • where does the fact that we compactify on a torus comes in? It is stated that the large amount of supersymmetry renders toroidal compactifications unrealistic. But I do not see where in the above discussion we specifically use the fact we compactify on a torus.

One more statement that follows a bit later is:

"Unbroken supersymmetry requires that the vacuum satisfies $\bar{\epsilon}Q |0\rangle$, where $\bar{\epsilon}(x^M)$ parametrizes the supersymmetry transformation which is generated by Q, and both Q and $\bar{\epsilon}$ are spinors of $SO(1, d_c-1)$."

I do not understand why this is the case.

Any help would be much appreciated!


1 Answer 1


The notation $\mathbf{n}$ for a number $n$ in the context of Lorentz representations denotes the dimensionality of this representation. E.g. denoting the Majorana representation of $\mathrm{SO}(10,1)$ as $\mathbf{32}$ is supposed to remind us that the vector space of such Majoranas is 32 dimensional. The reason to choose this notation is that then decomposing this in terms of smaller Lorentz representations of $\mathrm{SO}(3,1)\times\mathrm{SO}(7)$ is reminiscent of multiplication - note that $\mathbf{32} = (\mathbf{4},\mathbf{8})$ looks a lot like $32 = 4\cdot 8$.

For some reason, $(\mathbf{n}_1,\mathbf{n}_2)$ is the notation in this context for a tensor product - this is the representation $\mathbf{n}_1\otimes\mathbf{n}_2$ in proper mathematical parlance, and this is the reason the dimensionalities behave multiplicatively, since the dimension of a tensor product is the product of the dimensions of the multiplicands.

In order to see that $\mathbf{32} = (\mathbf{4},\mathbf{8})$ is actually true, just note that it must be a proper representation of $\mathrm{SO}(3,1)\times\mathrm{SO}(7)$ since that's just a subgroup of $\mathrm{SO}(10,1)$ and it is a representation of that, and that it cannot lose its Majorana character under reduction to the subgroup. There aren't any other representations except $\mathbf{4}$ and $\mathbf{8}$ that could really appear here. For non-Majorana representations it can be much more annoying to figure out what they decompose to.

(Yes, it is confusing that we do not attach to the $\mathbf{n}$ which group it is a representation of, after all, two representations of two different Lorentz groups can sometimes have the same dimensionality. No, people don't seem to be willing to switch to clearer notation. Get used to it.)

The $\mathcal{N}=8$ now comes from the $\mathbf{8}$ - remember that the seven compact dimensions are invisible from the viewpoint of the dimensionally reduced theory, and so all the 4d theory see is 8 copies of a 4d Majorana, it doesn't know anything about $\mathrm{SO}(7)$, it just sees the $\mathbf{8}$ as 8 completely independent dimensions. 8 copies means that the original single supercharge has become 8 supercharges in 4d, so we have $\mathcal{N}=8$.

Nothing in this logic cares about the torus - you haven't looked at anything for which the exact geometry of the compact dimensions would be relevant. You'll get to it - for instance, the geometry of the compactification manifold can influence the gauge groups in the resulting 4d theory.

  • $\begingroup$ Nice answer. But I don't see how this would hold in arbitrary geometries. In fact it sounds like the compactification they're really talking about is $S^6$ as this is the one that has an $SO(7)$ isometry group. $\endgroup$ Nov 9, 2021 at 18:10
  • $\begingroup$ @ConnorBehan The compactification manifold doesn't need to have $\mathrm{SO}(7)$ as a symmetry for what we're doing here - we can simply observe that there are eight $\mathbf{4}$ representations coming from the $\mathbf{32}$. The reduced theory doesn't care about the $\mathrm{SO}(7)$, after all. $\endgroup$
    – ACuriousMind
    Nov 9, 2021 at 22:46

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