Explicit Outer Automorphisms of Spi n(8)
1 Introduction The Lie groups Spi n(2n) have a nontrivial outer automorphism group. For n other than 4, this group is Z2 , which swaps the two spinor representations. One can see this by examining the Dynkin diagram, which for Spi n(2n) looks like ...
The two nodes on the right correspond to the spinor representations of Spi n(2n) and the nontrivial diagram automorphism, corresponding to the nontrivial outer automorphism of Spi n(2n), switches the two nodes. In the case of Spi n(8), the outer automorphism group is S 3 , the symmetric group on three elements. We can again see this from the Dynkin diagram:
which now has a full S 3 symmetry. The node on the far left of the diagram is the standard 8-dimensional rotation representation of SO(8), which we view here as lifted to a representation of Spi n(8). The two nodes on the right are still the spinor representations, which in this case are also 8-dimensional. We expect to thus be able to write down some automorphisms of matrices of Spi n(8) that are not inner automorphisms. To make this significantly easier, we look at the Lie algebra so(8, C). This avoids a few issues that crop up on the level of compact groups. So the aim is to now write down a basis for so(8, C) in which we can write non-inner automorphisms of so(8, C) without tearing our hair out.
2 Motivation for the construction Following the "four-ality" construction of Landsberg and Manivel [LM04], instead of identifying an S 3 action on so(8, C), we instead look for an S 4 action that yields an S 3 action. While there’s no S 4 action evident in the Dynkin diagram, we can find an S 4 action of a maximal subalgebra of so(8, C) and then extend it to the whole Lie algebra. 1
To find maximal subalgebras of so(8, C), we first look at the extended Dynkin diagram of so(8, C),
Removing a node yields the Dynkin diagram for a maximal subalgebra of so(8, C). If we remove the center node, we get
which corresponds to four commuting copies of sl (2, C), and this configuration has as clear S 4 action by permuting the four copies. We thus get that we can decompose so(8, C) as four copies of sl (2, C) and a 16dimensional vector space. Examining the roots of so(8, C) indicates that this 16dimensional vector space is just the tensor product of the 2-dimensional representations of the sl (2, C)s. So we get: so(8, C) = sl (2, C)1 ⊕ sl (2, C)2 ⊕ sl (2, C)3 ⊕ sl (2, C)4 ⊕ (V1 ⊗ V2 ⊗ V3 ⊗ V4 )
(1)
where sl (2, C)k acts on Vk . Thus S 4 acts on so(8, C) by permuting the copies of sl (2, C) and permuting the factors of V1 ⊗ V2 ⊗ V3 ⊗ V4 in the corresponding fashion. For the elements (12)(34), (13)(24) and (14)(23), the corresponding automorphisms is inner. Hence the outer automorphisms correspond to cosets of {i d , (12)(34), (13)(24), (14)(23)} within the S 4 , yielding the expected S 3 . The three 8-dimensional representations can be written in terms of the Vi : (V1 ⊗ V2 ) ⊕ (V3 ⊗ V4 ) (V1 ⊗ V3 ) ⊕ (V2 ⊗ V4 ) (V1 ⊗ V4 ) ⊕ (V2 ⊗ V3 ) We can see that (12)(34), (13)(24) and (14)(23) preserve each of these spaces. In the following, the matrices given will act on (V1 ⊗ V2 ) ⊕ (V3 ⊗ V4 ) in the usual fashion, so it is tempting to deem that the standard representation and the other two spinor representations. But the entire setup is symmetric so it doesn’t actually matter in any way.
3 The basis, abstractly We pick a basis {h, x, y} for sl (2, C) such that [h, x] = 2y, [h, y] = −2x and [x, y] = 2h. Then we take the 2-dimensional representation V of sl (2, C) and give it basis vectors e, f such that h.e = i e, h. f = −i f , x.e = − f and x. f = e. We apply this to all four copies of sl (2, C). We will write e 1 ⊗ e 2 ⊗ e 3 ⊗ e 4 as eeee and so on for the other basis vectors.
2
Starting from eeee, we can generate the other vectors by taking commutators with the x i in appropriate combinations. We pick the vector eeee such that £ ¤ eeee, f f f f £ ¤ eee f , f f f e £ ¤ ee f e, f f e f £ ¤ ee f f , f f ee £ ¤ e f ee, f e f f £ ¤ ef ef,f ef e £ ¤ e f f e, f ee f £ ¤ e f f f , f eee
=
i (h 1 + h 2 + h 3 + h 4 )
(2)
=
−i (h 1 + h 2 + h 3 − h 4 )
(3)
=
−i (h 1 + h 2 − h 3 + h 4 )
(4)
=
i (h 1 + h 2 − h 3 − h 4 )
(5)
=
−i (h 1 − h 2 + h 3 + h 4 )
(6)
=
i (h 1 − h 2 + h 3 − h 4 )
(7)
=
i (h 1 − h 2 − h 3 + h 4 )
(8)
=
−i (h 1 − h 2 − h 3 − h 4 )
(9)
The author would rather have had those all be positive i , but they can’t all have the same sign. The bases for the 8-dimensional representations can be written similarly in terms of e i and f i . In particular, we want eeee.ee00 = 0 and f f f f .ee00 to be some multiple of 00 f f , and so on for the other vectors. The S 4 action sending i to j sends h i to h j , x i to x j and y i to y j . The other sixteen basis vectors split into five orbits: two fixed points, two four-element orbits, and a six-element orbit.
4 The basis, concretely We can make an 8-by-8 matrix ·
so(4)12 −X T
X so(4)34
¸
where so(4)12 looks like 0 −h 1 − h 2 −x 1 − x 2 y1 − y2
h1 + h2 0 y1 + y2 x! − x2
x1 + x2 −y 1 − y 2 0 −h 1 + h 2
y2 − y1 −x 1 + x 2 h1 − h2 0
(10)
and similarly for so(4)34 . We can write so(4)12 acting on C4 with the following basis: 1 0 i 0 ee = 0 ,e f = 1 0 i
0 −1 0 i , f e = 1 , f f = 0 −i 0
where the signs are set by assuming that ee has the signs as listed and applying x 1 and x 2 in various combinations. 3
Then the X component of a basis element of V1 ⊗V2 ⊗V3 ⊗V4 can be written as 12 times the outer product of the corresponding pair of C4 vectors, with the 12-component being vertical and the 34-component being horizontal. So for example, ee f e becomes 1 0 0 1 −i £ ¤ 1 i 1 0 0 1 −i = 1 0 0 i (11) ee f e = 0 0 0 0 0 2 2 0
0
0
0
0
For the standard representation we use the same four vectors, ee, e f , f e, f f , now living in each component of C4 ⊕ C4 .
References LM04 Landsberg J. M. and Manivel L., Representation Theory and Projective Geometry, in Algebraic Transformation Groups and Algebraic Varieties, 71-122, Encyclopedia Math. Sci., 132, Springer 2004
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