derivations | Climbing Mount Bourbaki

I now want to talk about some of the material in Hartshorne, II.8. First, we need some preliminaries from commutative algebra.

Let ${A}$ be a commutative ring, ${B}$ an ${A}$ -algebra, and ${M}$ a ${B}$ -module. Then an ${A}$ -derivation of ${B}$ in ${M}$ is a linear map ${D: B \rightarrow M}$ satisfying ${D(a)=0}$ for ${a \in A}$ and

$\displaystyle D(bb') = (Db) b' + b (Db').$

The set of all such derivations forms a ${B}$ -module ${\mathrm{Der}_A(B,M)}$ . If we regard this as a set, clearly, we have a contravariant functor

$\displaystyle \mathrm{Der}_A(B, -): B-\mathrm{mod} \rightarrow \mathrm{Set}$

because if ${B \rightarrow B'}$ is a homomorphism of ${A}$ -algebras, we can pull back a derivation.

Before proceeding, I should say something about the canonical example. Let ${M}$ be a smooth manifold and ${O_x}$ the local ring (of germs of smooth functions) at ${x \in M}$ . Then ${\mathbb{R}}$ becomes an ${O_x}$ -module if the germ ${f}$ acts by multiplication by ${f(x)}$ . More precisely, we have an exact sequence

$\displaystyle 0 \rightarrow m_x \rightarrow O_x \rightarrow \mathbb{R} \rightarrow 0$

for ${m_x \subset O_x}$ the maximal ideal of functions vanishing at ${x}$ , and this is the way ${\mathbb{R}}$ is an ${O_x}$ -module.

Anyway, an ${\mathbb{R}}$ -derivation ${O_x \rightarrow \mathbb{R}}$ is just a tangent vector at ${x}$ .

Now back to the algebraic theory. It turns out that the functor ${\mathrm{Der}_A}$ is representable. In other words, for each ${A}$ -algebra ${B}$ , there is a ${B}$ -module ${\Omega_{B/A}}$ such that

$\displaystyle \hom_B( \Omega_{B/A}, M) \simeq \mathrm{Der}_A(B,M) ,$

the isomorphism being functorial. In addition, there must be a “universal” derivation ${d: B \rightarrow \Omega_{B/A}}$ (corresponding to the identity ${\Omega_{B/A} \rightarrow \Omega_{B/A}}$ in the above functorial isomorphism), that any derivation factors through.

The construction of ${\Omega_{B/A}}$ is straightforward. We define it as the ${B}$ -module generated by symbols ${db, b\in B}$ , modulo the relations ${da = 0}$ for ${a \in A}$ , ${d(b+b') = db + db'}$ , and ${d(bb') = b' db + b db'}$ . It is now clear that we have a functorial isomorphism as above. Now, ${\Omega_{B/A}}$ is called the module of Kahler differentials of ${B}$ over ${A}$ . (more…)

I learned the material in this post from the book by Humphreys on Lie algebras and representation theory.

Recall that if ${A}$ is any algebra (not necessarily associative), then the derivations of ${A}$ form a Lie algebra ${Der(A)}$ , and that if ${A}$ is actually a Lie algebra, then there is a homomorphism ${\mathrm{ad}: A \rightarrow Der(A)}$ . In this case, the image of ${\mathrm{ad}}$ is said to consist of inner derivations.

Theorem 1 Any derivation of a semisimple Lie algebra ${\mathfrak{g}}$ is inner.

To see this, consider ${\mathrm{ad}: \mathfrak{g} \rightarrow D :=Der(\mathfrak{g})}$ ; by semisimplicity this is an injection. Let the image be ${D_i}$ , the inner derivations. Next, I claim that ${[D, D_i] \subset D_i}$ . Indeed, if ${\delta \in D}$ and ${\mathrm{ad} x \in D_i}$ , we have

$\displaystyle [\delta, \mathrm{ad} x] y = \delta( [x,y]) - [x, \delta(y)] = [\delta(x),y] = (\mathrm{ad}(\delta(x)))y.$

In other words, ${[\delta, \mathrm{ad} x] = \mathrm{ad}(\delta(x))}$ . This proves the claim.

Consider the Killing form ${B_D}$ on ${D}$ and the Killing form ${B_{D_i}}$ on ${D_i}$ . The above claim and the definition as a trace shows that ${B_D|_{D_i \times D_i} = B_{D_i}}$ . (more…)

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Climbing Mount Bourbaki

Kahler differentials

Derivations of semisimple Lie algebras and the abstract Jordan decomposition

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