# 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

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

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

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

QED.

### 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 $C_G(\mathfrak{n}) = \{x \in G| Ad_x(\mathfrak{n} = \mathfrak{n}\}$.