Proof. At the Lie algebra level, we have the inclusion

so(4) so(6) ,! so(10)

by block diagonals, which is also just the di erential of the inclusion SO(4) SO(6) ,! SO(10) at the Lie group level. Given how the spinor reps are de ned in terms of creation and annihilation operators, it is easy to see that

commutes, because g is an intertwining operator between representations of so(4) so(6). That is because the so(4) part only acts on C2, while the so(6) part only acts on C3.

But these Lie algebras act by skew-adjoint operators, so really

commutes. Since the so(n)'s and their direct sums are semisimple, so are their images. Therefore, their images live in the semisimple part of the unitary Lie algebras, which is just another way of saying the special unitary Lie algebras. We get that

commutes, and this gives a commutative square in the world of simply connected Lie groups:

This result shows us how to reach the Spin(10) theory, not through the SU(5) theory, but through the Pati{Salam model. For physics texts that treat this issue, see for example Zee [40] and Ross [31].

3.5The Question of Compatibility

We now have two routes to the Spin(10) theory. In Section 3.2 we saw how to reach it via the SU(5) theory:

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/o /o /o/

More Uni cation

Our work in that section and in Section 3.1 showed that this diagram commutes, which is a way of saying that the Spin(10) theory extends the Standard Model.

In Section 3.4 we saw another route to the Spin(10) theory, which goes through Spin(4) Spin(6):

/o /o /o/

More Uni cation

Our work in that section and Section 3.3 showed that this diagram commutes as well. So, we have another way to extend the Standard Model and get the Spin(10) theory.

Drawing these two routes to Spin(10) together gives us a cube:

Are these two routes to Spin(10) theory the same? That is, does the cube commute?

Theorem 7. The cube commutes.

Proof. We have already seen in Sections 3.1-3.4 that the vertical faces commute. So, we are left with two questions involving the horizontal faces. First: does the top face of the cube

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commute? In other words: does a symmetry in GSM go to the same place in Spin(10) no matter how we take it there? And second: does the bottom face of the cube commute? In other words: does this triangle:

commute?

In fact they both do, and we can use our a rmative answer to the second question to settle the rst. As we remarked in Section 3.4, applying the map g to the Pati{Salam binary code given in Table 6, we get the SU(5) binary code given in Table 4. Thus, the linear maps f and gh agree on a basis, so this triangle commutes:

C2 C3

This in turn implies that the bottom face of the cube commutes, from which we see that the two maps from GSM to U( C5) going around the bottom face are equal:

The work of Section 3.1 through Section 3.4 showed that the vertical faces of the cube commute. We can thus conclude from diagrammatic reasoning that the two maps from GSM to U( C5) going around the top face are equal:

KKKKKKKKKK%

U( C5)

Since the Dirac spinor representation is faithful, the map Spin(10) ! U( C5) is injective. This means we can drop it from the above diagram, and the remaining square commutes. But this is exactly the top face of the cube. So, the proof is done. tu

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