# Subtori of groups of type E6

Let $$G$$ be a semisimple algebraic group of type $$E_6$$, defined over a perfect field $$k$$ (so $$G$$ is a group scheme over $$k$$ and $$G_{\bar{k}}$$ is a semisimple algebraic group in the usual sense), and let $$T$$ be a $$k$$-subtorus of $$G$$ of rank $$6$$, so a (not necessarily split) maximal subtorus of $$G$$.

Does there exist a strictly smaller semisimple subgroup $$H$$ of $$G$$ such that $$T$$ is a subtorus of $$H$$? What is the type of such $$H$$?

For example, can you find $$H$$ of type $$A_2\times A_2\times A_2$$ (probably too optimistic)?

• Is your main question whether it is true or false, that for every perfect field $k$, every form of $E_6$ over $k$, every maximal $k$-subtorus is contained in a proper semisimple $k$-subgroup? – YCor Oct 29 '18 at 9:25
• Yes, that was the first question, but I see now from a comment below that it can have negative answer. – Cehiju Oct 29 '18 at 16:42

## 1 Answer

This question is precisely answered by Borel–de Siebenthal theory. Ignoring rationality issues (i.e., base changing to an algebraic closure), and fundamental groups, the possible types of $$H$$ are $$E_6$$, $$A_1 + A_5$$, and $$A_2 + A_2 + A_2$$ (as you hoped, corresponding to removing the root $$\alpha_4$$ in Bourbaki's notation).

• For split $E_6$, all the split forms can be realised. Since Borel–de Siebenthal theory is constructive, it shouldn't be too hard to describe the other forms, but probably you'd want at least to specify some particular ground field to have any hope of, e.g., describing all possible tori in the first place. – LSpice Oct 29 '18 at 2:03
• Not exactly related, but possibly useful: garibaldibros.com/linked-files/almost-web.pdf – Victor Petrov Oct 29 '18 at 13:58
• Thank you @VictorPetrov. The propositions relative to $E^1_6$ and $E^2_6$ leave open the possibility that there might be subgroups of type $A_2^3$. Do you know if there is one case where only tori appear? – Cehiju Oct 29 '18 at 16:48
• For 3-special fields (that is whose finite extensions degrees are powers of 3) you always have a subgroup of type $A_2^3$, possibly of outer type (that is coming from a central simple algebra of degree 3 over a separable extension of degree 3). This follows from the fact that the order of the Weyl group of $E_6$ is divisible by $3^4$ and not by $3^5$. In general I don't know. – Victor Petrov Oct 30 '18 at 9:58