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  1. We denote by $F^{\mathbb{R}}_{p,q}$ the quadratic form over the field ${\mathbb{R}}$ $$ F^{\mathbb{R}}_{p,q}(x)=x_1^2+\dots+x_p^2-(x_{p+1}^2+\dots+x_{p+q}^2) $$ on the vector space $V^{\mathbb{R}}:={\mathbb{R}}^n$ (where $n=p+q$). By definition, $$ {\bf O}(p,q)={\rm Aut}({\mathbb{R}}^n,F^{\mathbb{R}}_{p,q}). $$ By Serre, Local Fields, Section X.2, Corollary 1 of Proposition 4, page 153 in the English version, there is a canonical bijection between $H^1({\mathbb{R}},{\bf O}(p,q))$ and the set of isomorphism classes of nondegenerate quadratic forms of $n$ variables over ${\mathbb{R}}$, i.e. isomorphism classes of pairs $(V,F)$ over ${\mathbb{R}}$ where $V$ is an $n$-dimensional vector space and $F$ is a nondegenerate quadratic form on $V$.

  2. Similarly, we denote by $F^{\mathbb{C}}_{p,q}$ the Hermitian ${\mathbb{C}}/{\mathbb{R}}$ form $$ F^{\mathbb{C}}_{p,q}(x)=x_1 \bar x_1+\dots+x_p\bar x_p-(x_{p+1}\bar x_{p+1}+\dots+x_{p+q}\bar x_{p+q}) $$ on the vector space $V^{\mathbb{C}}:={\mathbb{C}}^n$ (where $n=p+q$). By definition, $$ {\bf U}(p,q)={\rm Aut}({\mathbb{C}}^n,F^{\mathbb{C}}_{p,q}). $$ Then there is a canonical bijection between $H^1({\mathbb{R}},{\bf U}(p,q))$ and the set of isomorphism classes of nondegenerate Hermitian ${\mathbb{C}}/{\mathbb{R}}$ forms of $n$ variables, i.e. isomorphism classes of pairs $(V,F)$ over ${\mathbb{C}}$ where $V$ is an $n$-dimensional vector space and $F$ is a nondegenerate Hermitian form on $V$.

  3. Similarly we denote by $F^{\mathbb{H}}_{p,q}$ the Hermitian ${\mathbb{H}}/{\mathbb{R}}$ form on the vector space $V^{\mathbb{H}}:={\mathbb{H}}^n$ given by the same formula (where $n=p+q$ and by ${\mathbb{H}}$ we denote the skew field of Hamilton's quaternions with the canonical involution). By definition, $$ {\bf Sp}(p,q)={\rm Aut}({\mathbb{H}}^n,F^{\mathbb{H}}_{p,q}). $$ Then there is a canonical bijection between $H^1({\mathbb{R}},{\bf Sp}(p,q))$ and the set of isomorphism classes of nondegenerate Hermitian ${\mathbb{H}}/{\mathbb{R}}$ forms of $n$ variables.

Question 1. What is classified by $H^1({\mathbb{R}},{\bf SO}(p,q))$?

Question 2. What is classified by $H^1({\mathbb{R}},{\bf SU}(p,q))$?

I need explicit functorial descriptions as above rather than formulas for the cardinalities of the corresponding sets. My feeling is that in both questions $H^1$ classifies nondegenerate forms of n variables with the same determinant as $F_{p,q}$ (modulo squares in Question 1 and modulo norms in Question 2).

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  • $\begingroup$ The answer to Question 1 is: isomorphism classes of triples $(V,F,e)$ where $(V,F)$ is a non-degenerate quadratic space of rank $n$ over $\mathbf{R}$ and $e$ is a vector in $\det(V)$ on which the induced quadratic form has value $(-1)^{q}$; this $e$ is a kind of "orientation". Similar story for the other one. This works similarly over any field (not of char. 2) at all, replacing $(-1)^q$ with whatever value you like among those attained by the induced quadratic form on $\det(V)$ for the initially given space. $\endgroup$
    – user27920
    Commented Jul 3, 2014 at 15:26
  • $\begingroup$ @user52824: Thank you! As I expected, my questions reduce to linear algebra. Now what is the induced quadratic (resp., Hermitian) form on $\det(V)$? Could you please either explain or give references? I would be grateful if you could write an answer based on your comment. I will accept it immediately. $\endgroup$
    – Mikhail Borovoi
    Commented Jul 3, 2014 at 18:37
  • $\begingroup$ Using the dictionary between quadratic and symmetric bilinear (or hermitian and "conjugate-symmetric" semi-bilinear) forms, the issue of induced form on a tensor or exterior power becomes one for bilinear or semi-bilinear forms, which in turn is given by habitual formulas on elementary tensors or elementary wedge products. I don't know a reference since I figured these things out for myself when the need arose and I always assumed that everyone else did likewise; must be somewhere in Bourbaki (?). Try "tensor algebra, tensor pairings" in Google. $\endgroup$
    – user27920
    Commented Jul 3, 2014 at 19:09
  • $\begingroup$ By the way, this is all quite classical for Riemannian manifolds to equip the line bundle of top-degree differential forms with an induced metric tensor and thereby define in a coordinate-free manner the volume form associated to an orientation (as the unique fibral unit section in the half-line singled out by the orientation). $\endgroup$
    – user27920
    Commented Jul 3, 2014 at 19:12

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