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The isogenies of type $A_n$ are indexed by the subgroups of $\mathbb{Z} / (n+1) \mathbb{Z}$, i.e. by the positive divisors of $n+1$. If $a$ is a positive integer and $a \mid (n+1)$, then the cocharacter lattice corresponding to the isogeny is given by $\langle Q^{\vee}, \epsilon_a \rangle$, where $Q^{\vee}$ is the coroot lattice and $\epsilon_a$ is the fundamental coweight $$\epsilon_a = \frac{1}{n} [(n-a)(e_1 + e_2 + ... + e_a) - a(e_{a+1} + e_{a+2} + ... + e_{n+1}].$$

I would like a "nice" basis for this cocharacter lattice...nice for computational purposes (I am being vague for now on what this means). For example, if the group is simply connected, then (I would say that) the set of simple coroots form a nice basis. If the group is adjoint, then (I would say that) the set of fundamental coweights form a nice basis.

But what if the group is neither simply connected nor adjoint? Is there a standard/nice basis that people use when computing with these isogenies? Or is there a basis that you think seems "nice"?

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  • $\begingroup$ By "isogeny of type $A_n$", do you mean "integer lattice containing the root lattice of type $A_n$ with finite index"? $\endgroup$
    – LSpice
    Commented Jan 31, 2018 at 19:55
  • $\begingroup$ @LSpice: I mean lattice contained in the coweight lattice and containing the coroot lattice...in the setting of type $A_n$. $\endgroup$
    – Tippy Tipper
    Commented Jan 31, 2018 at 20:12
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    $\begingroup$ Note that the term "isogeny" has a technical meaning for algebraic groups, etc. Aside from this, it's useful to keep in mind that in type $A_n$, all fundamental weights $\varpi_i$ are minuscule, and moreover these are coset representatives (along with 0) for the $n+1$ cosets of the "fundamental group" (weight lattice modulo root lattice). Depending on the goal here, you might consider combining one of these weights with a basis of simple roots for the root lattice. These are not independent but might still be useful for computation. $\endgroup$
    – Jim Humphreys
    Commented Feb 1, 2018 at 0:04
  • $\begingroup$ @Jim Humphreys: Thanks for your comment. I think your suggestion is what I will end up pursuing. $\endgroup$
    – Tippy Tipper
    Commented Feb 1, 2018 at 13:34

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