Algebraic Groups II (lie algebra stuff)

As in the last post, G is an algebraic group.  There are at least two ways of defining the Lie(G) = \mathfrak{g}.  One depends on the left translation functions \lambda_x \colon k[G] \to k[G] defined via \lambda_x(f) = f(x^{-1}\cdot (\ )).  The other definition depends on defining the tangent space of G 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) \mathfrak{g} = left invariant vector fields on G = point derivations at e = T_e G = \mathfrak{g}.

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) Lie(G) = left invariant derivations on k[G] = \{d \in \mbox{Der}k[G]| d\circ \lambda_x = \lambda_x \circ d\}

The notation \mathfrak{g}, Lie(G) is to more explicitly differentiate between the definitions.  The correspondence between them is given by

Lie(G) \ni d \mapsto ev_e \circ d \in \mathfrak{g}

\mathfrak{g} \ni X \mapsto X\circ^* \lambda \in Lie(G)

where ev_e is the evaluation at the identity function and the derivation X\circ^* \lambda sends a function f to the function

g \mapsto X(\lambda_{g^{-1}}f).

This is plausible because \lambda_{g^{-1}}f \in k[G] and X \colon k[G] \to k; but it still an exercise to show that X\circ^* \lambda is actually a derivation.

To get a sense of the yoga that goes on in keeping these notions straight consider the adjoint action of G on its lie algebra.  In normal matrix land this described as Ad_g(X) = gXg^{-1} where everything is a matrix the juxtaposition is just matrix multiplication.  Compare in contrast

Prop: If d \in Lie(G) then Ad_g(d) = \rho_g \circ d \circ \rho_{g^{-1}} where \rho_gf = f((\ )\cdot x).

proof: Set \phi = Int_x \colon G \to G i.e. \phi(g) = xgx^{-1}.  Then d\phi_e = Ad_x.  Let \phi^\# \colon k[G] \to k[G] be the backwards map on on coordinate rings, then the map Ad_x \colon Lie(G) \to Lie(G) is Ad_x(d) = d \circ \phi^\#.  Note \phi^\#(f) = f(x(\ )x^{-1}) = \lambda_{x^{-1}} \circ \rho_{x^{-1}}f.  So \phi^\# \circ \rho_x = \lambda_{x^{-1}}.  By left invariance

d \circ \phi^\# \circ \rho_x = d \circ \lambda_{x^{-1}} = \lambda_{x^{-1}} \circ d

\Rightarrow ev_e \circ d \circ \phi^\# \circ \rho_x = ev_e \circ \lambda_{x^{-1}} \circ d = ev_e \circ \rho_x \circ d

\Rightarrow ev_e Ad_x(d) = ev_e \circ d \circ \phi^\# = ev_e \circ \rho_x \circ d \circ \rho_{x^{-1}}


Some Differentials

1. m \colon G \times G \to G; the (unsurprising) claim is that dm_e (X,Y) = X + Y

note m^\#(f) = (x,y \mapsto f(xy)) also can write m^\#(f) = \sum_i h_i\otimes g_i.  From f = f((\ )e) = f(e(\ )) it follows that f = \sum_i h_i g_i(e) = \sum_i h_i(e) g_i.

Note in general that T_e(Z \times W) = T_eZ \oplus T_eW:

Consider the point derivation \delta \in T_e(Z \times W) on f \otimes g \in k[Z]\otimes k[W].

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

= f(e)\delta(1\otimes g) + \delta(f \otimes 1)g(e)

so the map \delta \mapsto \bigr(\delta((\ )\otimes 1), \delta(1 \otimes (\ ))\bigr) gives the desired isomorphism.

So for (X,Y) \in \mathfrak{g} \oplus \mathfrak{g} and f \in k[G]

dm_e(X,Y).f = (X,Y).\sum_i h_i \otimes g_i

= \sum_i X(h_i)g_i(e) + h_i(e)Y(g_i)

= X.f + Y.f = (X + Y).f

2. i \colon G \to G is the inverse map.  Using that the composition G \xrightarrow{(1,i)} G\times G \xrightarrow{m} G is trivial (hence also on tangent spaces) its easily deduced that di_e(X) = -X.

3. Fix x \in G.  Set \gamma_x \colon G \to G by \gamma_x(y) = yxy^{-1}x^{-1}.  Set \phi_x(y) = yxy^{-1}.  Then d\phi_x = 1 - Ad_x.

From \gamma_x = m \circ (1, Int_x \circ i) it follows that

d\gamma_x = 1 - Ad_x

using the prop and and 1,2 and that Ad_x(-Y) = -Ad_x(Y).  Let c_x \colon G \to G be the constant map with image x.  Then \phi_x = m \circ (\gamma_x,x)  from which the claim follows.

Char 0 vs arbitrary characteristic

Thm: H \subset G is a closed, normal subgroup then \mathfrak{h} \subset \mathfrak{g} is an ideal.

proof: H normal \Rightarrow Int_x stabilizes H \Rightarrow Ad_x stabilizes \mathfrak{h}.  So if v_1,...,v_n is a basis of \mathfrak{g} and v_{n+1},..., v_m extends it to a basis of \mathfrak{g} then in this basis

Ad \colon G \to GL(\mathfrak{g})

x \mapsto \bigl(\begin{smallmatrix} * & * \\ 0 & * \end{smallmatrix} \bigr)

Then the differential ad \colon \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g}) also has this form.  One way to see this is that the image of a 1-parameter subgroup will look like

t \mapsto \bigl(\begin{smallmatrix} a(t) & b(t) \\ 0 & d(t) \end{smallmatrix} \bigr)

so differentiating and evaluating at t = 0 shows the claim.  Now apply this to x = \exp(Y) then we get ad_Y(\mathfrak{h}) \subset \mathfrak{h} for arbitrary Y \in \mathfrak{g}. QED.

Thm: if in addition char k = 0 and G,H are both connected then \mathfrak{h} is an ideal iff H is normal.

p oof: It remains to show H is normal when \mathfrak{h} \subset \mathfrak{g} is an ideal.  Consider the subgroup N = \{x \in G| Ad_g(\mathfrak{h}) \subset \mathfrak{h}\}.

this result depends on the following results

a) The association H \mapsto \mathfrak{h} gives a 1-1 inclusions preserving map

\{\mbox{closed conn. subgroups}\} \to \{\mbox{sub lie algebras}\}

b) Another powerful result implies that \mathfrak{n} = \{x \in \mathfrak{g}| ad_x(\mathfrak{h}) \subset \mathfrak{h}\}

Now b) \Rightarrow \mathfrak{g} = \mathfrak{n} \Rightarrow G = N.  Now consider Int_x \colon H \to xHx^{-1}; I have for every x \in G = N that the lie algebra of xHx^{-1} is Ad_x(\mathfrak{h}) = \mathfrak{h}.  So by a) H = xHx^{-1}.QED.

Thm G and algebraic group and x \in G then the lie algebra of C_G(x) = \{g|gxg^{-1}=x\} is contained in c_\mathfrak{g}(x) = \{Y|Ad_x(Y) = Y\}

Thm. If char k = 0 and G connected then the lie algebra of C_G(x) is exactly c_\mathfrak{g}(x).  Further \ker Ad = Z(G) and the lie algebra of \ker Ad is \ker ad = \mathfrak{z(g)}.

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

Thm: G is semisimple iff \mathfrak{g} is semisimple.

I’m not giving a proof but the intuition is to prove not one implies not the other.  For if G is not semismiple then it has a closed connected normal commutative subgroup N and then \mathfrak{n} is a commutative ideal in \mathfrak{g} so it’s not semisimple.  Going the other way, from a commutative ideal \mathfrak{n} \subset \mathfrak{g} consider C_G(\mathfrak{n}) = \{x \in G| Ad_x(\mathfrak{n} = \mathfrak{n}\}.


About this entry