2
$\begingroup$

Consider a simple Lie algebra $\mathfrak{g}$ and its Cartan subalgebra $\mathfrak{t}$ associated with a given Lie group $G$ with maximal torus $T$. In this paper we read below equation $(2.3)$ that $$T = \mathfrak{t}/\Lambda_{cochar.}$$ where $\Lambda_{cochar.} = \pi_1(T) = Hom(U(1),T)$ is the so-called "cocharacter lattice". Now, I do not understand very well the definition of this, e.g. why the fundamental group of the maximal torus is this lattice, but I want to understand the quotient.

The Cartan subalgebra $\mathfrak{t}$ is a vector space and by taking this quotient we take out some specific vectors. Ok. How on earth this is equal to the maximal torus? How something local as the Cartan subalgebra modulo some vectors is suddenly equal to some Riemannian manifold with generally some interesting geometry?

Is there some intuitive way to understand it? Say if we use $G = SL(2,\mathbb{C})$ then the maximal torus has rank one and the Cartan subalgebra is one-dimensional. How on earth the quotient of this Cartan subalgebra by some vectors gives me the maximal torus?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ You are looking at the quotient of a complex vector space by a lattice. This gives you a torus. For example, if $V = \Bbb C$ and $\Lambda = \Bbb Z^2$ you get $V/\Lambda \cong S^1 \times S^1$ is indeed a torus. $\endgroup$ – Nicolas Hemelsoet May 20 '18 at 18:36
  • $\begingroup$ I don't think you should think about such an identification as strict equality. There is a natural identification of each maximal torus with such a quotient, for reasons given above $\endgroup$ – leibnewtz May 20 '18 at 18:44
  • $\begingroup$ @NicolasHemelsoet why is this identified with the maximal torus though? $\endgroup$ – Gorbz May 20 '18 at 18:46
  • 1
    $\begingroup$ Compare this excellent answer, where the cocharacter lattice is called $\Lambda ^\vee$: math.stackexchange.com/a/1755110/96384 $\endgroup$ – Torsten Schoeneberg May 23 '18 at 6:06
1
$\begingroup$

A group morphism $u : S^1 \to T$ is called a cocharacter of $T$, and is naturally identified with $\pi_1(T) \cong \Bbb Z^n$.

Differentiating a cocharacter at $1$ gives you a map $ d_1u : \Bbb R \to \mathfrak t$. The image of $\Bbb Z$ by all the cocharacters is a lattice called the cocharacter lattice, written $\Lambda$ (it's easy to check that there is a canonical isomorphism with your definition). Integrating $d_1u$ with time $1$ gives you back $1 \in T$ (because $u$ is 1-periodic).

But for a general linear map $ \alpha : \Bbb R \to \mathfrak t$ there is no reason that $\alpha(1) \in \Lambda$. In fact we get an element $[\alpha] \in \mathfrak t/\Lambda$. In fact, two maps $v,v'$ gives the same element if and only if $v-v'$ is the derivative of a cocharacter i.e if $v(1) = v'(1) \text{ mod} \Lambda$

This gives a map $\mathfrak t/\Lambda \to T$ given by integration (i.e the exponential map from differential geometry), giving a canonical isomorphism between $T$ and $\mathfrak t/\Lambda$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks. Can you also provide a reference that the chocharacter lattice is explicitly mentioned? Also, $n$ in $\mathbb{Z}^n$ is the dimension of the maximal torus? $\endgroup$ – Gorbz May 20 '18 at 21:21
  • $\begingroup$ Yes $n= \dim T$ (it is called the rank of $G$). I am not sure about a reference. $\endgroup$ – Nicolas Hemelsoet May 20 '18 at 21:43
  • $\begingroup$ And in your 3rd paragraph how is $\alpha$ and $f$ related? $\endgroup$ – Gorbz May 21 '18 at 10:13
  • 1
    $\begingroup$ @Gorbz : it was a typo I edited. $\endgroup$ – Nicolas Hemelsoet May 21 '18 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.