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Chapter 8: Problem 22

Let \(A\) be any algebra over \(\mathbb{R}\). A derivation of \(A\) is a linear map \(D: A \rightarrow A\) satisfying \(D(x y)=(D x) y+x(D y)\) for all \(x, y \in A\). Show that if \(D_{1}\) and \(D_{2}\) are derivations of \(A\), then \(\left[D_{1}, D_{2}\right]=D_{1} \circ D_{2}-D_{2} \circ D_{1}\) is also a derivation. Show that the set of derivations of \(A\) is a Lie algebra with this bracket operation.

Short Answer

Expert verified
The commutator \([D_1, D_2]\) is a derivation, and derivations form a Lie algebra with bracket operation.

Step by step solution

01

Understand the task

The problem requires us to prove that a particular operation on derivations, called the commutator (or Lie bracket), results in another derivation. Then we show that the set of all derivations forms a Lie algebra.
02

Recall the definition of a derivation

A derivation on an algebra \(A\) is a linear map \(D: A \rightarrow A\) satisfying the Leibniz rule: \(D(xy) = (Dx)y + x(Dy)\) for all \(x, y \in A\).
03

Describe the commutator of two derivations

Given two derivations \(D_1\) and \(D_2\), the commutator \([D_1, D_2]\) is defined as \([D_1, D_2] = D_1 \circ D_2 - D_2 \circ D_1\).
04

Prove that the commutator is a derivation

We need to show that \([D_1, D_2](xy) = ([D_1, D_2]x)y + x([D_1, D_2]y)\). Compute \([D_1, D_2](xy) = D_1(D_2(xy)) - D_2(D_1(xy))\), and apply the Leibniz rule to both terms.
05

Apply the Leibniz rule to \(D_2(xy)\) and \(D_1(xy)\)

For \(D_2(xy)\), use \(D_2(xy) = (D_2x)y + x(D_2y)\) and for \(D_1(xy)\), use \(D_1(xy) = (D_1x)y + x(D_1y)\). Then substitute these into the expression obtained in Step 4.
06

Simplify the expression

After substitution, we have: \([D_1, D_2](xy) = (D_1(D_2x))y + x(D_1(D_2y)) - (D_2(D_1x))y - x(D_2(D_1y))\). Reorganize the terms to match the form \(([D_1, D_2]x)y + x([D_1, D_2]y)\).
07

Conclusion for derivation

Since the expression from Step 6 matches the form required by the Leibniz rule, \([D_1, D_2]\) is a derivation.
08

Define Lie algebra

A Lie algebra is a vector space equipped with a bilinear operation (here, the commutator) satisfying anticommutativity and the Jacobi identity.
09

Verify anticommutativity

Verify that \([D_1, D_2] + [D_2, D_1] = 0\), which follows from the commutator's definition: \([D_1, D_2] = -(D_2 \circ D_1 - D_1 \circ D_2)\).
10

Verify the Jacobi identity

For any derivations \(D_1, D_2, D_3\), the Jacobi identity is \([D_1, [D_2, D_3]] + [D_2, [D_3, D_1]] + [D_3, [D_1, D_2]] = 0\). This can be verified through substitution and simplification.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivation in Algebra
In the context of algebra, a derivation is a specific kind of linear map. Imagine you have an algebraic structure, say an algebra \( A \), and you want to understand how elements within interact under this map. A derivation, \( D: A \to A \), is defined by a particular rule known as the Leibniz rule. This rule states that for any elements \( x \) and \( y \) in the algebra:
  • \( D(xy) = (Dx)y + x(Dy) \)
This rule is crucial because it preserves a product structure even under linear transformations.
Think of a derivation like a differential operator that "spreads out" across a product of functions, similar to how the derivative works in calculus. It retains the fundamental attributes of multiplication within a given set by ensuring it behaves predictably under the derivation operation.
Commutator
The commutator is a fascinating operation when you're working with derivations. Given two derivations \( D_1 \) and \( D_2 \), the commutator is defined as:
  • \([D_1, D_2] = D_1 \circ D_2 - D_2 \circ D_1 \)
The idea behind the commutator is to measure the extent to which these derivations fail to commute, or in simpler terms, how much order matters in their operation. If \( [D_1, D_2] = 0 \), this means the operations commute, offering potential symmetry or simpler structures in certain scenarios.
This operation is not arbitrary: knowing how components like \( D_1 \) and \( D_2 \) interact helps better understand the structure of the algebra itself. More importantly, the result of this operation, the new map \([D_1, D_2]\), is also a derivation. This has profound implications, helping us organize derivations into a structured system known as a Lie algebra.
Leibniz Rule
The Leibniz rule is a cornerstone of derivation theory in algebra, analogous to the product rule in calculus. It prescribes how a derivation should interact with the product of two elements. Formally, for any two elements \( x \) and \( y \) in an algebra \( A \), and any derivation \( D \), the rule is:
  • \( D(xy) = (Dx)y + x(Dy) \)
This rule ensures that the "derivation" operator respects the algebraic structure it acts on. It guarantees that applying \( D \) to a product distributes through in a specific way, mirroring the distributive property familiar in arithmetic operations.
In proving that a given operation, such as the commutator of derivations, is itself a derivation, the Leibniz rule becomes an essential tool. By applying this rule, we affirm that the structure and behavior intended by a derivation are maintained even while dealing with complex combinations or transformations of derivations.

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Most popular questions from this chapter

Suppose \(F: G \rightarrow H\) is a Lie group homomorphism. Show that the kernel of \(F_{*}: \operatorname{Lie}(G) \rightarrow \operatorname{Lie}(H)\) is the Lie algebra of Ker \(F\) (under the identification of the Lie algebra of a subgroup with a Lie subalgebra as in Theorem 8.46).

Suppose \(G\) is a Lie group and \(g\) is its Lie algebra. A vector field \(X \in \mathscr{X}(G)\) is said to be right-invariant if it is invariant under all right translations. (a) Show that the set \(\bar{g}\) of right-invariant vector fields on \(G\) is a Lie subalgebra of \(\mathfrak{X}(G)\). (b) Let \(i: G \rightarrow G\) denote the inversion map \(i(g)=g^{-1}\). Show that the pushforward \(i_{*}: \mathfrak{X}(G) \rightarrow \mathfrak{X}(G)\) restricts to a Lie algebra isomorphism from \(\mathrm{g}\) to \(\overline{\mathrm{g}}\).

Suppose \(F: M \rightarrow N\) is a smooth submersion, where \(M\) and \(N\) are positivedimensional smooth manifolds. Given \(X \in \mathfrak{X}(M)\) and \(Y \in \mathfrak{X}(N)\), we say that \(X\) is a lift of \(Y\) if \(X\) and \(Y\) are \(F\)-related. A vector field \(V \in \mathfrak{X}(M)\) is said to be vertical if \(V\) is everywhere tangent to the fibers of \(F\) (or, equivalently, if \(V\) is \(F\)-related to the zero vector field on \(N\) ). (a) Show that if \(\operatorname{dim} M=\operatorname{dim} N\), then every smooth vector field on \(N\) has a unique lift. (b) Show that if \(\operatorname{dim} M \neq \operatorname{dim} N\), then every smooth vector field on \(N\) has a lift, but that it is not unique. (c) Assume in addition that \(F\) is surjective. Given \(X \in \mathfrak{X}(M)\), show that \(X\) is a lift of a smooth vector field on \(N\) if and only if \(d F_{p}\left(X_{p}\right)=d F_{q}\left(X_{q}\right)\) whenever \(F(p)=F(q)\). Show that if this is the case, then \(X\) is a lift of a unique smooth vector field. (d) Assume in addition that \(F\) is surjective with connected fibers. Show that a vector field \(X \in \mathfrak{X}(M)\) is a lift of a smooth vector field on \(N\) if and only if \([V, X]\) is vertical whenever \(V \in \mathscr{X}(M)\) is vertical.

(a) Given Lie algebras \(g\) and \(h\), show that the direct sum \(g \oplus h\) is a Lie algebra with the bracket defined by $$ \left[(X, Y),\left(X^{\prime}, Y^{\prime}\right)\right]=\left(\left[X, X^{\prime}\right],\left[Y, Y^{\prime}\right]\right) \text {. } $$ (b) Suppose \(G\) and \(H\) are Lie groups. Prove that \(\operatorname{Lie}(G \times H)\) is isomorphic to \(\operatorname{Lie}(G) \oplus \operatorname{Lie}(H)\).

Let \(M\) be a smooth manifold with or without boundary, let \(N\) be a smooth manifold, and let \(f: M \rightarrow N\) be a smooth map. Define \(F: M \rightarrow M \times N\) by \(F(x)=(x, f(x))\). Show that for every \(X \in \mathfrak{X}(M)\), there is a smooth vector field on \(M \times N\) that is \(F\)-related to \(X\).
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