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I apologize in advance if my question is too elementary for MO.

It is a well known fact that the linear algebraic group $G = \mathsf{SO}_n$ is connected, and there exist a few different proofs of this fact. One proof goes by showing that $G$ is generated by unipotent elements, and invoking the theorem that every linear algebraic group with this property is connected.

My question is about a different, more direct proof, involving the Cayley transform $$ A \mapsto (I_n + A)^{-1} (I_n - A) , $$ which maps every matrix $A \in G(k)$ for which $I_n + A$ is invertible (let's write $W$ for the set of such matrices), to a skew-symmetric matrix, and in fact, this map defines an isomorphism of varieties between the non-empty open subset $W$ of $G$, and an open subset of the irreducible variety $V$ of skew-symmetric matrices.

But how does one now conclude that $\mathsf{SO}_n$ is connected? In particular, why is the closure of $W$ equal to $G$? Of course the complement of $W$ is given by the polynomial equation $\det(I_n + A) = 0$, but how is this situation different from, for instance, the fact that $\mathsf{SO}_n$ is the subvariety of $\mathsf{O}_n$ defined by the polynomial equation $\det(A) - 1 = 0$ (or $\det(A) + 1 \neq 0$) whereas $\mathsf{O}_n$ is not connected?

What subtlety am I missing?

added:

  1. I'm assuming $\operatorname{char}(k) \neq 2$.
  2. The underlying topology is the Zariski toplogy.
  3. The proof I'm mentioning in the second paragraph is only valid in characteristic $0$.
  4. It seems that the question is not too elementary...

added: I'll add a few more sentences to make clear what I mean by connected in the Zariski topology.

I'm considering $G$ as a group functor (i.e. a functor from the category of commutative $k$-algebras to the category of groups). Then $G$ is connected if and only if its coordinate algebra $k[G] = \mathcal{O}(G)$ has no non-trivial idempotents. In particular, in order to show that $G$ is connected, we may assume w.l.o.g. that $k$ is algebraically closed. [In fact, since $G$ is smooth (since $\operatorname{char}(k) \neq 2$), we may instead consider the group of $\bar{k}$-rational points $G(\bar{k})$, which brings us back to a more classical situation.]

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    $\begingroup$ I am a little confused by the question: Skew matrices form an affine space, and your set of nonsingular matrices is an affine hypersurface (which is presumably easy to show is irreducible). $O_n$ is a much more complicated variety, so this Cayley transform trick seems to have simplified your life. $\endgroup$ – Igor Rivin Jun 5 '12 at 17:54
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    $\begingroup$ Presumably the OP means connected as a scheme. $\endgroup$ – Igor Rivin Jun 5 '12 at 20:10
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    $\begingroup$ I'm slightly confused by your second paragraph: the finite group $\mathbf{Z}/p\mathbf{Z}$ may be viewed as a linear algebraic group. Over a field of char. p, it is generated by unipotent elements. But it is not connected. $\endgroup$ – George McNinch Jun 5 '12 at 23:39
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    $\begingroup$ @Robert, George: I think Tom was starting out in characteristic 0, which is one reason for my attempted clarifications below. $\endgroup$ – Jim Humphreys Jun 6 '12 at 0:28
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    $\begingroup$ @Tom: I have to confess that I'm still puzzled. It appears that you are only avoiding characteristic $2$, so that you would appear to be claiming that $\mathsf{SO}_2(k)$ is connected in the Zariski topology even when $k$ is finite, and I don't understand this. After all, when $k=\mathbb{Z}_3$, doesn't ${\mathsf{SO}}_2(k)$ have $4$ elements and isn't the complement of any element open? Why then is this group not disconnected? Looking at the answers below, it seems that you at least need $k$ to be infinite for those kinds of arguments to work. $\endgroup$ – Robert Bryant Jun 6 '12 at 18:53
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Each connected component of an algebraic group has the same dimension. Thus if it has a connected subvariety whose complement has a lower dimension, it is connected. (If it had two connected components, only one could contain the connected subvariety, and so the other would have a lower dimension.)

The subvariety where $I_n+A$ is invertible is birational to $\mathbb A^{\frac{n(n-1)}{2}}$ and thus has dimension $\frac{n(n-1)}{2}$.

The complement is where $A$ has a $-1$ eigenvalue. But since every eigenvalue to an orthogonal matrix must have its inverse also an eigenvalue, the determinant is the product of all the eigenvalues which are their own inverse, which are just $-1$ and $1$. Thus, if the determinant is $1$, the number of $-1$ eigenvalues is even, so is at least $2$. We can split the matrix into a $2$-dimensional $-1$-eigenspace and a matrix in $SO(n-2)$. These things are paramaterized by a $SO_{n-2}$-bundle on $G_{2}^{n-2}$, whose dimension is $2(n-2)+\frac{(n-2)(n-3)}{2}=\frac{n(n-1)}{2}-1$. If the $-1$-eigenspace is more than two-dimensional there is more than one way to express a matrix in this way, but that can only decrease the dimension.

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  • $\begingroup$ Thanks a lot! The dimension argument is indeed what I was missing. I'm relieved to see that the answer is not as obvious as I feared. $\endgroup$ – Tom De Medts Jun 6 '12 at 8:15
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    $\begingroup$ I think this solution is incomplete, in particular it isn't true that we can always split the matrix into a 2-dim $-1$-eigenspace and a matrix in $SO(n-2)$ (at least the way I interpret this statement, that we have an orthogonal decomposition into invariant subspaces). For an example, take $g = -u$ where $u$ is a regular unipotent element in $SO_4$. It's not hard to see that there's no 2-dimensional subspace (let alone one on which $g = -1$) that is non-degenerate and $g$-invariant. $\endgroup$ – fherzig Feb 9 '17 at 22:53
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    $\begingroup$ @fherzig Good point! I think there is a way to fix this, where you split into cases. There must be an eigenvalue -1. If it has nonzero norm then you can split it off, and there must be another -1 eigenvalue. If it has nonzero norm, you can split both off and my argument is valid. Otherwise, there must be a -1 eigenvalue with nonzero norm. Consider the filtration into that one-dimensional subspace and its $n-1$ dimensional perpendicular space. There are $(n-2)(n-3)/2 + (n-2)$ degrees of freedom in the action on that, plus $n-2$ degrees of freedom in choosing a one-dimensional $-1$ eigenspace. $\endgroup$ – Will Sawin Feb 10 '17 at 6:41
  • $\begingroup$ @fherzig The count is because, in block-diagonal form, there is an $n-2 \times n-2$ $SO_{n-2}$ block, two $n-2 \times 1$ blocks that determine each other, and a $1 \times 1$ block fixed by the other stuff. $\endgroup$ – Will Sawin Feb 10 '17 at 6:42
  • $\begingroup$ @Will: that's nice, thanks! (btw, in the "otherwise" sentence you meant "zero norm".) $\endgroup$ – fherzig Feb 11 '17 at 16:50
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[EDIT: I've tightened my wording and added a couple of references which I went back to out of curiosity.]

Will's answer has the elements needed for a concrete reply to the question, but the question itself has caused some confusion about the setting and terminology which are worth clarifying. First of all, the underlying field should be of characteristic different from 2, since it gets more subtle to talk about quadratic forms and orthogonal groups in characteristic 2. (This is done however in work of Chevalley and Borel in the algebraic groups framework, where the groups are included in the classification.)

Originally the study of orthogonal groups as Lie groups was carried out by Weyl, Chevalley, and many others. Here the (polynomial) condition on $n \times n$ matrices is just that the transpose of a matrix must equal its inverse. The orthogonal matrices then form a compact real Lie group $\mathrm{O}(n)$ or a noncompact complex Lie group $\mathrm{O}(n, \mathbb{C})$ of dimension $n(n-1)/2$. In the euclidean topology, the latter group is homeomorphic to the former group times a vector space. So connectedness questions can be settled in the compact case.

Since eigenvalues of an orthogonal matrix occur along with their inverses, $\det=\pm 1$ and matrices of det $-1$ form a closed normal subgroup $\mathrm{SO}(n)$ or $\mathrm{SO}(n, \mathbb{C})$ giving in Lie theory the rank $\ell$ series: $B_\ell$ with $n=2\ell+1$ odd, $D_\ell$ with $n=2\ell$ even. It's worth following the case $n=5$ in Will's calculations. To show that the compact group is connected in the topological group setting, Chevalley in Theory of Lie Groups uses induction on $n$ and the characterization of the successive quotients as spheres.

Now in the Chevalley-Borel setting of linear algebraic groups (over an algebraically closed field $K$), much of the previous study carries over with modifications. For linear algebraic groups given the Zariski topology, irreducibility of the underlying variety fortunately coincides with connectedness in that coarse topology; the term "connected" is then preferred. The irreducible (= connected) components of an algebraic group $G$ are disjoint and equidimensional as well as finite in number (unlike some Lie groups): these are just the cosets of the identity component $G^\circ$. We denote the points of the group over $K$ as $\mathrm{SO}_n(K)$, but the scheme language probably adds nothing useful to the study of connectedness here.

The most standard elementary way to show that a linear algebraic group is connected is to show that it is generated by suitable irreducible subsets such as closed connected subgroups. For the classical matrix groups, this is usually done by showing that the group is generated by transvections, hence by connected 1-dimensional unipotent groups. With some care this approach even handles special orthogonal groups in characteristic 2.

On the other hand, the question here raises the possibility of appealing (in characteristic not 2) to a Cayley transform. Here one is able to map isomorphically a nonempty open subset of an affine space (dense in the Zariski topology) onto a nonempty open subset of the matrix group in a concrete way. Then it has to be seen, as Will shows, that none of the hypothetical extra irreducible/connected components of $\mathrm{SO}_n(K)$ can lie in the excluded hypersurface given by nonvanishing of a determinant. Dimension counting seems necessary here.

The only source I can quote for this slightly esoteric approach is a terse exercise 2.2.2(2) in Springer's book Linear Algebraic Groups, where much is left to the reader's ingenuity. (Are there earlier sources?) Springer himself was attracted to this approach, I think, because he used the Cayley transform for classical groups to realize an isomorphism between unipotent and nilpotent varieties in the group and its Lie algebra.

Earlier arguments appear in at least two places. [Note in each case that for the standard structure theory (over an arbitrary field) involving an isotropic split torus in diagonal form, orthogonal groups are written as matrices using an orthogonal direct sum of hyperbolic planes; over $K$ this translates to the conventional format above.]

1) Chevalley's 1956-58 seminar Classification des groupes algebriques semi-simples (typeset text, Springer 2005, edited by Cartier). In Expose 22, Chevalley gives an argument for connectedness of some of the linear groups roughly analogous to the inductive argument for compact Lie groups.

2) The second edition of Borel's original notes Linear Algebraic Groups (Springer GTM 126, 1991). In the added Section 23 he discusses examples involving groups of rational points of various classical groups, observing in particulqr that over $K$ the relevant groups are Zariski-connected. (Characteristic 2 requires as usual extra discussion, as does type $D_\ell$.) Here the argument relies on the standard structure theory, showing in effect that a hypothetical coset representative for $G/G^\circ$ must in fact represent an element of the Weyl group and thus lie in $G^\circ$.

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    $\begingroup$ What a great answer! I wish I could accept more than one answer in MO, since the combination of Will's answer and yours is marvelous. One of the sources where I learned about the Cayley transform was indeed Springer's Exercise 2.2.2(2), and the fact that it appeared as an exercise was the main reason for my apologetic first paragraph. $\endgroup$ – Tom De Medts Jun 6 '12 at 8:14
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This is a more elementary solution, which would make sense in the context of Springer's book. We consider $SO(V,\langle,\rangle)$, or for short $SO(V)$, where $V$ is $n$-dimensional and $\langle,\rangle$ a nondegenerate symmetric bilinear form on $V$. (Recall that any two such are equivalent.)

We'll show by induction on $n$ that $SO(V)$ is connected. Springer's exercise shows that any $g \in SO(V)$ that doesn't have eigenvalue $-1$ is in the identity component of $SO(V)$.

When $n = 1$, $SO(V) = 1$ and when $n = 2$ we see directly that $SO(V)$ is connected. (In fact, $SO_2 = \{\left(\begin{matrix} a & b\\ -b & a\end{matrix}\right) : a^2 + b^2 = 1\}$ and $SO_2 \cong \mathbb G_m$ as algebraic groups via $\left(\begin{matrix} a & b\\ -b & a\end{matrix}\right) \mapsto a+bi$ with $i^2 = -1$.)

In general, fix $g \in SO(V)$ and we'll show that $g$ is contained in the identity component. Note that $V = \oplus V_\lambda$ is a direct sum of generalised eigenspaces and that $V_\lambda \perp V_\mu$ if $\lambda \ne \mu^{-1}$. Thus we get an orthogonal decomposition $V = V_{-1} \perp V_1 \perp \perp_{\{\lambda \ne \lambda^{-1}\}} (V_\lambda \oplus V_{\lambda^{-1}})$. In particular, $V_{-1}$ is a nondegenerate subspace and we have the orthogonal decomposition of $g$-invariant subspaces $V = V_{-1} \perp V_{-1}^\perp$. Also, we get that $\dim V_\lambda = \dim V_{\lambda^{-1}}$ if $\lambda \ne \lambda^{-1}$, so by considering the determinant we see that $\dim V_{-1}$ even (as remarked by Will), say it equals $2d$.

It follows that $g$ is contained in the subgroup $SO(V_{-1}) \times SO(V_{-1}^\perp)$ of $SO(V)$. By induction we are thus reduced to the two extreme cases: $V = V_{-1}$ or $V = V_{-1}^\perp$. The second case is treated by the Springer exercise, so we may assume that all eigenvalues of $g$ are $-1$ and $n = 2d$.

Now $-g$ is unipotent, hence by the Springer exercise we know it is in the identity component of $SO(V)$. Multiplying by $-1 \in SO(V)$ we deduce that $g$ is in the same component as $-1$. Finally, after choosing an appropriate basis $-1$ lies in the subgroup $SO_2 \times \dots \times SO_2$ of $SO_{2d}$, so we are done by the $n = 2$ case.

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  • $\begingroup$ Incidentally, the same argument works for $Sp_{2n}$. Again one has a Cayley transform $Sp_{2n} --> Lie Sp_{2n}$ given by the same formula as for the orthogonal case. This time one uses that $Sp_2 = SL_2$ is connected. $\endgroup$ – fherzig Feb 10 '17 at 2:16
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    $\begingroup$ If one treats the connectedness of Sp$_{2m}$ via an inductive fibration method in the spirit of what is done for ${\rm{SL}}_N$ then there is the modest gain that one gets the connectedness for Sp$_{2m}$ in characteristic 2 as well, which is very helpful for the inductive proof of connectedness of SO$_n$ in characteristic 2 for $n$ of both parities. $\endgroup$ – nfdc23 Feb 10 '17 at 4:57

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