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I have been reading Cassels's book on "Rational Quadratic Forms". Most part of his book is written perfectly, but there is a chapter about the "Spin Representation" on his book, which I can not really understand, basically because I don't have enough motivation why such a thing would be important.

So I want to understand, the motivation behind the spin representation, and more importantly some application of the concept.

Presumably this notion came from the Geometry, and I guess this is why, for example, they prove spin group is a double cover of proper orthogonal group of a quadratic form. Which kind of result one might obtain by considering this geometric interpolation in the context of quadratic forms?

Thank you.

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    $\begingroup$ I do not see a chapter with that title. $\endgroup$
    – Will Jagy
    Commented Dec 21, 2011 at 5:13
  • $\begingroup$ Will@: Look at chapter 10, 11. $\endgroup$
    – M.B
    Commented Dec 21, 2011 at 5:16
  • $\begingroup$ chapter 10 is called The Spin and Orthogonal Groups. Chapter 11 is called Spinor Genera. I do not see your chapter. $\endgroup$
    – Will Jagy
    Commented Dec 21, 2011 at 5:17

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It was a longstanding problem to decide equivalence of indefinite forms. The showpiece of the spinor genus is that, for indefinite forms in at least three variables over the rational integers, the spinor genus and the equivalence class coincide.

The phenomena that are most directly explained occur in three variables and positive forms, first identified in Jones and Pall (1939). The first example is that $$ g(x,y,z) = 2 x^2 + 2 y^2 + 5 z^2 + 2 y z + 2 z x \neq m^2,$$ where all prime factors of $m$ are congruent to $1 \pmod 4.$ In comparison, and in the same genus, $$ f(x,y,z) = x^2 + y^2 + 16 z^2$$ does represent all squares, and primitively represents all $m^2.$ Other than the difference noted, the two forms represent the same numbers. $g$ and $f$ are in different spinor genera. $f$ is called regular, as it represents all numbers locally eligible, while $g$ is called spinor regular, as it represents everything eligible for its spinor genus.

Kaplansky, A. Schiemann, and I found all possible regular positive ternary forms in 1997. Later, in contact with Andrew Earnest, i found a total of 29 spinor regular forms that are not regular. This is probably the complete list. Earnest contacted me recently about completing the project by proving the 29 are all. I sent him some background on what is necessary to complete such a proof.

My own little toy, to appear next year, is infinite families of genera where membership in a spinor genus can be decide by a single number. If $N$ is squarefree and $N = u^2 + v^2$ in integers, then consider the genus of $$ h(x,y,z) = x^2 + y^2 + 16 N z^2. $$ It turns out that the genus of $h$ has exactly two spinor genera, there are spinor exceptional integers, which are all the $Nm^2.$ The theorem, quite unusual, is that a form $k(x,y,z)$ in the same genus as $h$ is in the same spinor genus as $h$ if and only if $k$ integrally represents $N.$ This is pretty rare. There is usually no assurance about the small numbers a form represents. All we really know is that a form resents all sufficiently large numbers that are primitively represented by some form in the same spinor genus; this is Duke and Schulze-Pillot (1990). I have about a dozen such infinite families and some related conjectures.

Probably enough for now.

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  • $\begingroup$ Thanks a lot for your comment. It seems very interesting. My initial reason for asking this question, was more computational. Would you please tell me why the spin group of $q(x,y,z)=xy+z^2$ is essentially $SL_2(\mathbb{Z})$? and do we have a nice algorithm to compute the spin group? $\endgroup$
    – M.B
    Commented Dec 21, 2011 at 6:12
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I believe the historical motivation for considering the spin representation is actually mathematical physics. In his attempt to formulate a counterpart of the Schrodinger equation in quantum mechanics which is compatible with special relativity, Dirac decided that he needed to find a first order differential operator whose square is the Laplacian:

$$ D = \sum c_i \partial_i $$ $$ D^2 = -\sum \partial_i^2 $$

If you sit down and work out the relations that the $c_i$'s have to satisfy, you find that $c_i c_j + c_j c_i = 0$ if $i \neq j$ and $c_i^2 = -1$. Of course, there are no real or complex numbers which satisfy these relations; Dirac realized (not in this language) that the $c_i$'s are generators for the Clifford algebra $\mathbb{C}_n$. Thus $D$ is a vector valued operator which takes values in a vector space $S$ on which $\mathbb{C}_n$ acts by endomorphisms. Ideally we would like to take $S$ as small as possible, in the sense that $End(S) \cong \mathbb{C}_n$.

If this physicsy motivation isn't enough, Atiyah and Singer discovered a few decades later that Dirac's idea has profound applications in differential topology. One says that a Riemannian $n$-manifold $M$ is spin if there is a principal $Spin(n)$-bundle $P \to M$ equipped with a bundle map $P \to SO(TM)$ (where $SO(TM)$ is the principal $SO(n)$-bundle of orthonormal frames) which on fibers is the double cover $Spin(n) \to SO(n)$. If $P$ exists then there is a natural vector bundle $S \to M$ coming from the spin representation via the "associated bundle" construction, and $S$ has the property that its endomorphism bundle is isomorphic to the Clifford algebra bundle associated with $TM$. In this case there is a global "Dirac operator" $D$ whose square is the Laplacian, and this operator is central to the Atiyah-Singer index theorem. In fact, it serves as the fundamental class in real K-homology theory.

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    $\begingroup$ Small nitpick: surely you mean $ c_i c_j + c_j c_i =0 $ Also, Dirac actually wanted to find such an operator whose square was the D'Alembert operator, the Laplacian in space-time, so with one +-sign (the time direction) and three minus-signs (space directions), so the $c_i$ generate the Clifford algebra $Cl_{3,1}$ (or $Cl_{1,3}$, I always forget the signs). $\endgroup$ Commented Dec 21, 2011 at 13:25
  • $\begingroup$ Quite right - thanks for the corrections! I edited my answer to account for the former. $\endgroup$ Commented Dec 22, 2011 at 3:23
  • $\begingroup$ Though I must say, I always wondered, why you want to have S as small as possible. In every book it just says you do (and you inserted this phrase in your exposé as well), but I never really understood why this would be important. Especially since some Manifolds don't admit a spin structure, but you can always look at the bundle of Clifford algebras and the Dirac operator living there. $\endgroup$ Commented Dec 22, 2011 at 8:49
  • $\begingroup$ @Kofi: The short answer is that $S$ is a direct summand of any complex representation of the Clifford algebra, so the spinor bundle . The long answer is more subtle. The basic issue is that the index of an operator is determined not just by the operator itself but by whatever "super-symmetric" data happens to be lying around. Specifically a local ONB $\{e_i\}$ acts by right multiplication on the bundle of Clifford algebras as supersymmetries (odd, self-adjoint, squares to $-1$), and these supersymmetries anti-commute with the Dirac operator. $\endgroup$ Commented Dec 22, 2011 at 19:24
  • $\begingroup$ A bundle on $M^n$ which is locally isomorphic to the Clifford algebra $\mathbb{C}_n$ which also comes equipped with local supersymmetries as above is the same thing as the choice of a spin$^c$ structure on $M$, and in the even dimensional case one can use the supersymmetries to reduce the Clifford algebra down to the ordinary spin representation. So your question comes down to: why must we remember the supersymmetry data? Generally the answer is that if you forget them you end up solving a trivial index problem. This is most transparent in the K-homology proof. $\endgroup$ Commented Dec 22, 2011 at 19:35

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