# Is $({F^{\times})^{ diag}}\backslash(GL_2 \times E^{\times})_{det=\mathbb{N}}$ a unitary group?

Let $$F$$ be an p-adic field, and $$E$$ be a quadratic extension of $$F$$, then is $$({F^{\times})^{ diag}}\backslash(GL_2 \times E^{\times})_{det=\mathbb{N}}$$ isomorphic to some unitary group $$U_{E/F}(2)$$? Here, $$\mathbb{N}(e)=e *\sigma(e)$$, where $$\sigma$$ is the nontrivial element in $$Gal(E/F)$$. And $$(GL_2 \times E^{\times})_{det=\mathbb{N}} :=\{(g,e)\in GL_2 \times E^{\times}|det(g)=\mathbb{N}(e)\}$$.

• Can you explain the notation ${\rm det}={\mathbb N}$? – Paul Broussous Jan 8 at 18:13
• Does "det=\mathbb N" mean that the determinant is a norm, by any chance? – paul garrett Jan 8 at 22:33
• Also, I think $U_{E/F}(2)$ is not reliably universal notation. I'm guessing that you mean the hermitian form to be two diagonal 1's, and conjugation is Galois conjugation? – paul garrett Jan 8 at 22:34
• I have edited the problem again. – Cooler Panda Jan 9 at 16:09
• The answer depends on whether you regard $({F^{\times})^{ diag}}\backslash(GL_2 \times E^{\times})_{det=\mathbb{N}}$ as an algebraic group over $F$ or as a topological group (the group of points). – Mikhail Borovoi Jan 9 at 20:58

I cannot answer this question for now. However, I have made some relevant calculations that are too long for a comment.

We set $$G_0=(GL_2 \times E^{\times})_{{\rm det}={\rm Nm}}\,,\quad G_1=G_0/(F^\times)^{\rm diag},\quad G_2=U_{E/F}(1,-\lambda),$$ where $$\lambda\in F^\times$$ and $$U_{E/F}(1,-\lambda)$$ denotes the unitary group of the diagonal hermitian form $$h(z_1,z_2)={\rm Nm}(z_1)-\lambda{\rm Nm}(z_2).$$

We wish to know whether $$G_1$$ and $$G_2$$ are isomorphic. If $$\lambda\notin{\rm im}({\rm Nm})$$, then the hermitian form $$h$$ does not represent 0, the topological group $$G_2$$ is compact, and hence, it is not isomorphic to the noncompact group $$G_1$$ (this has been already noticed by Paul Broussous). Therefore, let us assume that $$\lambda\in{\rm im}({\rm Nm})$$; then we may assume that $$\lambda=1$$ and $$G_2=U_{E/F}(1,-1)$$.

Consider the homomorphism $$\alpha\colon G_0\to F^\times,\quad (g,e)\mapsto {\rm Nm}(e).$$ Clearly, $${\rm im}(\alpha)={\rm im}({\rm Nm})\subset F^\times$$. The homomorphism $$\alpha$$ induces a homomorphism $$\beta\colon G_1\to F^\times/F^{\times\,2},$$ and we have $${\rm im}(\beta)= {\rm im}({\rm Nm})/F^{\times\,2}.$$ We set $$U_1=\{e\in E\ |\ {\rm Nm}(e)=1\}.$$

We have a canonical homomorphism $$\gamma\colon{\rm SL}(2,F)\times U_1\to G_1\quad (g,e)\mapsto [g,e],$$ where $$[g,e]$$ denotes the class of the pair $$(g,e)\in G_0$$. Then $$\ker(\gamma)=\{\pm 1\},\quad {\rm im}(\gamma)=\ker(\beta).$$ Thus we obtain a short exact sequence $$1\to ({\rm SL}(2,F)\times U_1)/\{\pm 1\}\to G_1\to {\rm im}({\rm Nm})/F^{\times\,2}\to 1.$$

On the other hand, we have a canonical homomorphism $$\delta\colon {{\rm SU}_{E/F}}(1,-1)\times U_1\to G_2 \quad (s,e)\mapsto se,$$ whose kernel is $$\{\pm 1\}$$ and whose cokernel is $$U_1/U_1^2$$, where we write $$U_1^2$$ for the subgroup of squares in $$U_1$$. Thus we obtain a short exact sequence $$1\to({{\rm SU}_{E/F}}(1,-1)\times U_1)/\{\pm 1\}\to G_2\to U_1/U_1^2\to 1.$$ We have $${{\rm SU}_{E/F}}(1,-1)\simeq{\rm SL}(2,F)$$ (see, for instance, the book on classical groups by Jean Dieudonné or the book Geometric Algebra by Emil Artin). Thus we obtain a short exact sequence $$1\to({\rm SL}(2,F)\times U_1)/\{\pm 1\}\to G_2\to U_1/U_1^2\to 1.$$

Comparing the short exact sequences for $$G_1$$ and for $$G_2$$, we see that if the 2-groups $${\rm im}({\rm Nm})/F^{\times\,2}$$ and $$U_1/U_1^2$$ are non-isomorphic (that is, have different orders), then $$G_1$$ and $$G_2$$ are not isomorphic "in a nice way".

I do not know whether the 2-groups $${\rm im}({\rm Nm})/F^{\times\,2}$$ and $$U_1/U_1^2$$ are isomorphic or not. I suggest for OP to ask this on MathOverflow.