Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I've been using the fact that Spin(3,2) is isomorphic to Sp(4, R) for a while, but I've never seen a proof. Can anyone point me in the direction of a good reference?

share|improve this question
add comment

2 Answers 2

Here is a coordinate-free version. Let $(V, \psi)$ be a 4-dimensional symplectic space over a field, and let $\omega \in (\wedge^2 V)^{\ast} = \wedge^2(V^{\ast})$ be the nonzero 2-form arising from $\psi$. Then on the 6-dimensional vector space $W = \wedge^2(V)$ the kernel of $\omega$ is a hyperplane $H$ of dimension 5, and on $W$ there is a natural non-degenerate quadratic form $q$ valued in the line $\wedge^4(V)$ via $q(w) = (1/2)(w \wedge w)$ for $w \in W$ (e.g., if $w = e_1 \wedge e_2 + e_3 \wedge e_4$ for a basis $\{e_i\}$ of $W$ then $q(w) = e_1 \wedge e_2 \wedge e_3 \wedge e_4$); the definition of $q$ uses base change from the $\mathbf{Z}_{(2)}$-flat case for $(V,\psi)$ if $2$ isn't a unit.

By computing in linear coordinates of $V$ that "standardize" $\psi$ we see that $q$ is a split quadratic form on $W$, and the action of ${\rm{SL}}(V)$ on $W$ clearly preserves $q$ while the action of its subgroup ${\rm{Sp}}(V,\psi)$ preserves $H$. Hence, the action on $H$ defines a map $${\rm{Sp}}(V,\psi) \rightarrow {\rm{O}}(q|_H),$$ so this lands inside ${\rm{SO}}(q|_H)$ and as such defines a homomorphism $${\rm{Sp}}_4 = {\rm{Sp}}(V,\psi) \rightarrow {\rm{SO}}(q|_H) = {\rm{SO}}_5.$$

This map kills the center $\mu_2$ inside ${\rm{Sp}}_4$, and thereby identifies ${\rm{Sp}}_4$ as the degree-2 "simply connected" central cover of ${\rm{SO}}_5$ (in the sense of algebraic groups). Hence, this uniquely lifts to an isomorphism of ${\rm{Sp}}_4$ onto ${\rm{Spin}}_5$.

A nice feature of this conceptual construction is that it kills two birds with one stone: if we don't restrict to $H$ and instead work with the entire 6-dimensional $W$ then a similar construction defines the isomorphism of ${\rm{SL}}_4 = {\rm{SL}}(V)$ onto ${\rm{Spin}}_6$.

share|improve this answer
    
To avoid confusion, I think you should perhaps write $\operatorname{SO}_{3,2}$ and $\operatorname{Spin}_{3,2}$ for $\operatorname{SO}_5$ and $\operatorname{Spin}_5$, respectively, and similarly, $\operatorname{Spin}_{3,3}$ for $\operatorname{Spin}_6$ in the last line. –  José Figueroa-O'Farrill Jul 15 '13 at 10:20
2  
@Jose: I am using standard notation among algebraists who study algebraic groups over an arbitrary field $k$, where non-degenerate quadratic forms are not classified by a "signature", but typically a lot more data. One writes ${\rm{SO}}(q)$ and ${\rm{Spin}}(q)$ for the connected semisimple $k$-groups associated to a non-degenerate quadratic space $(V,q)$, and when $q$ is split with $V$ of rank $n$ then they are denoted ${\rm{SO}}_n$ and ${\rm{Spin}}_n$. For $k=\mathbf{R}$, one writes ${\rm{SO}}(r,s)$ to denote ${\rm{SO}}(q)(k)$ for $q$ of signature $(r,s)$. So I prefer to leave it as is. –  user36938 Jul 15 '13 at 12:36
    
I do agree it's good to learn how to avoid "choice-of-coordinates" silliness. But/and some of the intrinsification is better understood after one has done things an ad-hoc way. (I think we all know this, although it is not high-rep to admit it.) Then there is the field-specific issue of "classification" of (non-degenerate) quadratic forms over a particular field, which is surely not solvable in general... But, over any algebraically closed field of char not 2 (such as complex...) it's just dimension, and a slightly non-trivial result: over $\mathbb R$, it's "signature". Not "just algebra". –  paul garrett Jul 15 '13 at 23:55
    
@paul: even in char. 2 everything works perfectly well based on char-free definitions, as it must if there is to be a reasonable theory over $\mathbf{Z}$ for linking it up with Chevalley groups. The "correct" dichotomy for SO and Spin groups is not char. 2 versus char. $\ne 2$ but rather $\dim V$ being even or odd (i.e., B versus D). –  user36938 Jul 16 '13 at 1:19
    
Perhaps quadratic forms experts are less uneasy about Arf invariants than I am... :) –  paul garrett Jul 16 '13 at 16:38
show 4 more comments

By coincidence, I put up a protraced working-out of several of these examples today, at http://www.math.umn.edu/~garrett/m/v/sporadic_isogenies.pdf

Edit: Over algebraically closed fields (especially in char not $2$, which I want to not think about), these sporadic isogenies are easy to write down. In principle, but I think not in practice for most of us, to see what happens over not algebraically closed fields is a question of "Galois cohomology", as in Weil's "Algebras and classical groups" paper. For me, it's much easier to use a few coordinates. E.g., although $SU(4)\rightarrow SO(6)$ and $SU(2,2)\rightarrow SO(4,2)$, apparently $SU(3,1)$ does not map to $SO(p,q)$ with $p+q=6$. Meanwhile, $SL_2(\mathbb H)\rightarrow SO(5,1)$. Seems weird to me.

share|improve this answer
    
Perfect. Thanks! –  MRD1729 Jul 14 '13 at 20:42
2  
For completeness: Paul lets $Sp_4(R)$ act on the 16-dimensional matrix algebra $M_4(R)$ by conjugation. The usual bilinear form $tr(xy)$ is invariant. The Lie algebra $RI_4+sp_4(R)$ is an invariant subspace on which this form is nondegenerate ($sp_4(R)$ being the 10-dimensional Lie algebra), so its orthogonal is also nondegenerate, it is thus (16-(10+1))-dimensional, i.e. 5-dimensional. Explicit coordinates show the signature is (3,2). This defines a 2-to-1 map from $Sp_4(R)$ onto $SO(3,2)$. –  YCor Jul 14 '13 at 21:57
    
@YvesCornulier, thx for filling in! :) –  paul garrett Jul 15 '13 at 23:51
    
Great, that seems to be very practical! –  The User Jul 16 '13 at 10:57
    
@paulgarrett: $SU(3,1)$ that you ask about maps to $SO^*(6)$ (a.k.a. $SO(3,\mathbb H)$), according to Helgason (§X.6.4) or Besse (Einstein Manifolds, p. 201). You may need to add a third section "Over $\mathbb H$"! –  Francois Ziegler Jul 23 '13 at 4:30
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.