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An algebraic group over a field $K$ is called a split algebraic group if it has a Borel subgroup that has a composition series such that all the composition factors are isomorphic to either the additive group of $K$ or the multiplicative group of $K$. Are there some examples of non-split algebraic groups? Thank you very much.

LJR
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    There are lots of examples in number theory, especially in the Langlands program and class field theory. The group of units in a central simple algebra over a number field is non-split at some places. Quaternion algebras arise in this way, for instance. – davidlowryduda Sep 11 '15 at 02:35
  • @mixedmath surely you do not need to invoke class field theory or Langlands (!) to give examples of non-split groups :) – guest Sep 11 '15 at 02:50

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Take any non-split (connected) torus $T/k$. Clearly $T$ itself is the only Borel in $T$ and the composition series is just $T$ itself. Thus, $T$ is split if and only if $T\cong\mathbf{G}_m^k$. But there are lots of examples of tori not isomorphic to the multiplicative group. The most canonical examples I can think of are:

  • The Deligne torus $\mathbb{S}:=\mathrm{Res}_{\mathbb{C}/\mathbb{R}}\mathbf{G}_{m,\mathbb{C}}$. This classifies Hodge structures—more precisely, it's the algebraic group associated to the neutral Tannakian category of Hodge structures over $\mathbb{R}$.
  • The unit norm elements of $\mathrm{Res}_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}$. These are the smooth locus of elliptic curves with non-split multiplicative reduction.
  • The unitary group $U(1)/\mathbb{R}$.

In general, one can classify the non-split tori over $K$ as the non-trivial elements of

$$H^1(G_K,\mathrm{GL}_n(\mathbb{Z}))=\mathrm{Hom}_\text{cont.}(G_K,\mathrm{GL}_n(\mathbb{Z}))$$

Or, said differently, the tori over $K$ correspond to finitely generated free $\mathbb{Z}$-modules with a continuous action of $G_K$ (for the discrete topology on the group).

This makes the '$\mathrm{Res}$' above seem more obvious. Namely, if we have a finite extension $L/K$ then what's a natural way of going from a $G_L$-representation $\rho:G_L\to\mathrm{GL}_n(\mathbb{Z})$ to a representation $\rho:G_K\to\mathrm{GL}_n(\mathbb{Z})$? By induction! Namely, if $M$ is our free $\mathbb{Z}$-representation of $G_L$ then $\mathrm{Ind}_{G_L}^{G_K} M$ gives a representation of $G_K$. If the torus corresponding to $M$ is $T$ then the torus corresponding to $\mathrm{Ind}_{G_L}^{G_K}M$ is $\mathrm{Res}_{L/K}T$.

EDIT: The torus $U(1)/\mathbb{R}$ is given by

$$U(1)=\left\{\begin{pmatrix}a & b\\ -b & a\end{pmatrix}:a^2+b^2=1\right\}\subseteq\mathrm{GL}_{2,\mathbb{R}}$$

This is not isomorphic to $\mathbf{G}_{m,\mathbb{R}}$. For instance, $U(1)(\mathbb{R})$ is just the unit circle $S^1$ which has lots of torsion elements, but $\mathbb{R}^\times$ has only two torsion elements. That said, $U(1)$ is a commutative connected algebraic group, and so is its only Borel is itself. This shows that $U(1)$ is not split in your definition.

I leave it as an exercise to check that $U(1)_\mathbb{C}\cong\mathbf{G}_{m,\mathbb{C}}$ (let me know if you any difficulties).

Alex Youcis
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    With no disrespect, if the OP can process the examples you are posting, it is unlikely they would be asking this question. I think it is best to present the example of $U(1)$ over $\mathbb{R}$ and demonstrate how it is non-split and how it becomes split after extension to $\mathbb{C}$. – guest Sep 11 '15 at 02:57
  • @guest Thank you for the input. I'll add an edit! – Alex Youcis Sep 11 '15 at 03:02
  • By the way, you should not delete any of your remarks, since other more advanced users can benefit. However, for the newcomer, I think an emphasis on the real and complex forms of $U(1)$ is the easiest and most illuminating way to understand the split/non-split dichotomy. – guest Sep 11 '15 at 03:05
  • Now I think the answer is perfect. I already upvoted. – guest Sep 11 '15 at 04:43
  • @guest Thank you! I think you helped improve my answer markedly. :) – Alex Youcis Sep 11 '15 at 04:49
  • @Alex Youcis, thank you very much. What is the meaning of the notation ${\bf G}{m, \mathbb{R}}$? Is ${\bf G}{m, \mathbb{R}} = \mathbb{R}^{\times}$? Is $U(1)_{\mathbb{C}} = {a \in \mathbb{C}: |a|=1}$? – LJR Sep 12 '15 at 01:59
  • @LJR The notation $\mathbf{G}{m,\mathbb{R}}$ is the mulitiplicative group over $\mathbb{R}$. Also, note quite. What you wrote down are the $\mathbb{R}$-points of $U(1)/\mathbb{R})$. In fact, as I said above, $U(1)\mathbb{C}$ is the multiplicative group over $\mathbb{C}$. – Alex Youcis Sep 12 '15 at 02:01
  • @Alex Youcis, thank you very much. By definition, what is $U(1)_{\mathbb{C}}$? – LJR Sep 12 '15 at 02:06
  • @Alex Youcis, how to show that $U(1){\mathbb{C}} = {\bf G}{m, \mathbb{C}}$? – LJR Sep 12 '15 at 02:24