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Added text on the questioner's specific group, and reference to Chernousov's result in Platonov & Rapinchuk
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You have to be kind of careful. For the complex numbers, there is no problem as BCnrd wrote above. For a general field $K$, you are asking a basic question from the field of geometric algebra (there's an AMS GTM volume by Grove on the topic, and Dieudonne's books on classical groups).

If you take a nondegenerate quadratic form $q$, then we know:

  • If $q$ is isotropic over $K$, then $Spin(q)(K)$ is "projectively simple" (i.e., the quotient by its finite center is simple)
  • If $q$ is anisotropic, then $Spin(q)(K)$ can be far from simple. You can construct examples using valuations, as you suggest.

But you asked about $SO(q)(K)$. Then you use Galois cohomology (where I assume that $K$ has characteristic different from 2 for simplicity):

$1 \to \mu_2(K) \to Spin(q)(K) \to SO(q)(K) \to K^{\times}/K^{\times 2}$

where the last map is the spinor norm. You should think of the spinor norm as usually having a big image, so $SO(q)(K)$ will be pretty far from simple.

Your specific group

You asked specifically about $SO(q)$ where $q$ is a sum of squares. Then the spinor norm map has image products of sums of squares (typically not actually squares themselves), so you should expect the image of $Spin(q)(K)$ to be a normal subgroup in $SO(q)(K)$ of large index. (That is all very imprecise, but a precise answer depends on the arithmetic of $K$ and the dimension of $q$.) But now you can ask: Is $Spin(q)(K)$ modulo its center a simple group?

Here you have a strong advantage. If $q$ is isotropic over $K$ (i.e., you can write $0$ as a sum of a small enough number of nonzero squares), then you know from classical results that the answer is "yes".

If $q$ is not isotropic over $K$, then $Spin(q)$ is anisotropic (as an algebraic group) but it is split by the quadratic extension $K(\sqrt{-1})$. For such groups, under some hypotheses on $K$ (maybe as weak as characteristic zero), you know that the answer is again "yes" by Chernousov. See Theorem 9.7 on page 514 of the book "Algebra & Number Theory" by Platonov and Rapinchuk. You will have to inspect the proof (which starts on p.546) to see the precise hypotheses they need on $K$, or you can consult some of the original papers, where the proof is slightly different.

You have to be kind of careful. For the complex numbers, there is no problem as BCnrd wrote above. For a general field $K$, you are asking a basic question from the field of geometric algebra (there's an AMS GTM volume by Grove on the topic, and Dieudonne's books on classical groups).

If you take a nondegenerate quadratic form $q$, then we know:

  • If $q$ is isotropic over $K$, then $Spin(q)(K)$ is "projectively simple" (i.e., the quotient by its finite center is simple)
  • If $q$ is anisotropic, then $Spin(q)(K)$ can be far from simple. You can construct examples using valuations, as you suggest.

But you asked about $SO(q)(K)$. Then you use Galois cohomology (where I assume that $K$ has characteristic different from 2 for simplicity):

$1 \to \mu_2(K) \to Spin(q)(K) \to SO(q)(K) \to K^{\times}/K^{\times 2}$

where the last map is the spinor norm. You should think of the spinor norm as usually having a big image, so $SO(q)(K)$ will be pretty far from simple.

You have to be kind of careful. For the complex numbers, there is no problem as BCnrd wrote above. For a general field $K$, you are asking a basic question from the field of geometric algebra (there's an AMS GTM volume by Grove on the topic, and Dieudonne's books on classical groups).

If you take a nondegenerate quadratic form $q$, then we know:

  • If $q$ is isotropic over $K$, then $Spin(q)(K)$ is "projectively simple" (i.e., the quotient by its finite center is simple)
  • If $q$ is anisotropic, then $Spin(q)(K)$ can be far from simple. You can construct examples using valuations, as you suggest.

But you asked about $SO(q)(K)$. Then you use Galois cohomology (where I assume that $K$ has characteristic different from 2 for simplicity):

$1 \to \mu_2(K) \to Spin(q)(K) \to SO(q)(K) \to K^{\times}/K^{\times 2}$

where the last map is the spinor norm. You should think of the spinor norm as usually having a big image, so $SO(q)(K)$ will be pretty far from simple.

Your specific group

You asked specifically about $SO(q)$ where $q$ is a sum of squares. Then the spinor norm map has image products of sums of squares (typically not actually squares themselves), so you should expect the image of $Spin(q)(K)$ to be a normal subgroup in $SO(q)(K)$ of large index. (That is all very imprecise, but a precise answer depends on the arithmetic of $K$ and the dimension of $q$.) But now you can ask: Is $Spin(q)(K)$ modulo its center a simple group?

Here you have a strong advantage. If $q$ is isotropic over $K$ (i.e., you can write $0$ as a sum of a small enough number of nonzero squares), then you know from classical results that the answer is "yes".

If $q$ is not isotropic over $K$, then $Spin(q)$ is anisotropic (as an algebraic group) but it is split by the quadratic extension $K(\sqrt{-1})$. For such groups, under some hypotheses on $K$ (maybe as weak as characteristic zero), you know that the answer is again "yes" by Chernousov. See Theorem 9.7 on page 514 of the book "Algebra & Number Theory" by Platonov and Rapinchuk. You will have to inspect the proof (which starts on p.546) to see the precise hypotheses they need on $K$, or you can consult some of the original papers, where the proof is slightly different.

Source Link
Skip
Skip
  • 2.2k
  • 17
  • 23

You have to be kind of careful. For the complex numbers, there is no problem as BCnrd wrote above. For a general field $K$, you are asking a basic question from the field of geometric algebra (there's an AMS GTM volume by Grove on the topic, and Dieudonne's books on classical groups).

If you take a nondegenerate quadratic form $q$, then we know:

  • If $q$ is isotropic over $K$, then $Spin(q)(K)$ is "projectively simple" (i.e., the quotient by its finite center is simple)
  • If $q$ is anisotropic, then $Spin(q)(K)$ can be far from simple. You can construct examples using valuations, as you suggest.

But you asked about $SO(q)(K)$. Then you use Galois cohomology (where I assume that $K$ has characteristic different from 2 for simplicity):

$1 \to \mu_2(K) \to Spin(q)(K) \to SO(q)(K) \to K^{\times}/K^{\times 2}$

where the last map is the spinor norm. You should think of the spinor norm as usually having a big image, so $SO(q)(K)$ will be pretty far from simple.