# Algebraic Groups II (lie algebra stuff)

As in the last post, is an algebraic group. There are at least two ways of defining the . One depends on the left translation functions defined via . The other definition depends on defining the tangent space of at the identity and in turn there are at least three ways to think about tangent spaces; here I’ll think of the tangent space as point derivations.

(vector fields) left invariant vector fields on = point derivations at = .

The whole point of saying ‘left invariance’ is that such vector fields are determined by their value at the identity, so it seems superfluous to think of it as vector fields, but then again it connects well with the next definitions)

(derivations) left invariant derivations on =

The notation is to more explicitly differentiate between the definitions. The correspondence between them is given by

where is the evaluation at the identity function and the derivation sends a function to the function

.

This is plausible because and ; but it still an exercise to show that is actually a derivation.

To get a sense of the yoga that goes on in keeping these notions straight consider the adjoint action of on its lie algebra. In normal matrix land this described as where everything is a matrix the juxtaposition is just matrix multiplication. Compare in contrast

Prop: If then where .

proof: Set i.e. . Then . Let be the backwards map on on coordinate rings, then the map is . Note . So . By left invariance

QED.

### Some Differentials

1. ; the (unsurprising) claim is that

note also can write . From it follows that .

Note in general that :

Consider the point derivation on .

latex \delta(f\otimes 1 \cdot 1 \otimes g)$

so the map gives the desired isomorphism.

…

So for and

2. is the inverse map. Using that the composition is trivial (hence also on tangent spaces) its easily deduced that .

3. Fix . Set by . Set . Then .

From it follows that

using the prop and and 1,2 and that . Let be the constant map with image . Then from which the claim follows.

### Char 0 vs arbitrary characteristic

Thm: is a closed, normal subgroup then is an ideal.

proof: normal stabilizes stabilizes . So if is a basis of and extends it to a basis of then in this basis

Then the differential also has this form. One way to see this is that the image of a 1-parameter subgroup will look like

so differentiating and evaluating at shows the claim. Now apply this to then we get for arbitrary . QED.

Thm: if in addition char and are both connected then is an ideal iff is normal.

p oof: It remains to show is normal when is an ideal. Consider the subgroup .

this result depends on the following results

a) The association gives a 1-1 inclusions preserving map

b) Another powerful result implies that

Now b) . Now consider ; I have for every that the lie algebra of is . So by a) .QED.

Thm and algebraic group and then the lie algebra of is contained in

Thm. If char and connected then the lie algebra of is exactly . Further and the lie algebra of is .

This last result is just for characteristic zero. Let be a connected an algebraic group; it’s semisimple if it has no nontrivial closed connected normal commutative subgroups.

Thm: is semisimple iff is semisimple.

I’m not giving a proof but the intuition is to prove not one implies not the other. For if is not semismiple then it has a closed connected normal commutative subgroup and then is a commutative ideal in so it’s not semisimple. Going the other way, from a commutative ideal consider .

## About this entry

You’re currently reading “Algebraic Groups II (lie algebra stuff),” an entry on Math Meandering

- Published:
- July 4, 2010 / 3:03 pm

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- alg. geo.

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