Jordan–Chevalley decomposition
In mathematics,[] the Jordan–Chevalley decomposition, named after Camille Jordan and Claude Chevalley, expresses a linear operator as the sum of its commuting semisimple part and its nilpotent parts. The multiplicative decomposition expresses an invertible operator as the product of its commuting semisimple and unipotent parts. The decomposition is easy to describe when the Jordan normal form of the operator is given, but it exists under weaker hypotheses than the existence of a Jordan normal form. Analogues of the Jordan-Chevalley decomposition exist for elements of linear algebraic groups, Lie algebras, and Lie groups, and the decomposition is an important tool in the study of these objects.
Decomposition of a linear operator
Consider linear operators on a finite-dimensional vector space over a field. An operator T is semisimple if every T-invariant subspace has a complementary T-invariant subspace. An operator x is nilpotent if some power xm of it is the zero operator. An operator x is unipotent if x − 1 is nilpotent.Now, let x be any operator. A Jordan–Chevalley decomposition of x is an expression of it as a sum
where xs is semisimple, xn is nilpotent, and xs and xn commute. Over a perfect field, such a decomposition exists, the decomposition is unique, and the xs and xn are polynomials in x with no constant terms. In particular, for any such decomposition over a perfect field, an operator that commutes with x also commutes with xs and xn.
If x is an invertible operator, then a multiplicative Jordan–Chevalley decomposition expresses x as a product
where xs is semisimple, xu is unipotent, and xs and xu commute. Again, over a perfect field, such a decomposition exists, the decomposition is unique, and xs and xu are polynomials in x. The multiplicative version of the decomposition follows from the additive one since, as is easily seen to be invertible,
and is unipotent.
If x is written in Jordan normal form then xs is the endomorphism whose matrix contains just the diagonal terms of x, and xn is the endomorphism whose matrix contains just the off-diagonal terms; xu is the endomorphism whose matrix is obtained from the Jordan normal form by dividing all entries of each Jordan block by its diagonal element.
Proof of uniqueness and existence
The uniqueness follows from the fact are polynomial in x: if is another decomposition such that and commute, then, and both commute with x, hence with . Now, the sum of commuting semisimple endomorphisms is again semisimple. Since the only operator which is both semisimple and nilpotent is the zero operator it follows that and.We show the existence. Let V be a finite-dimensional vector space over a perfect field k and an endomorphism.
First assume the base field k is algebraically closed. Then the vector space V has the direct sum decomposition where each is the kernel of, the generalized eigenspace and x stabilizes, meaning. Now, define so that, on each, it is the scalar multiplication by. Note that, in terms of a basis respecting the direct sum decomposition, is a diagonal matrix; hence, it is a semisimple endomorphism. Since is then whose -th power is zero, we also have that is nilpotent, establishing the existence of the decomposition.
The fact that are polynomials in x follows from the Chinese remainder theorem. Indeed,
let be the characteristic polynomial of x. Then it is the product of the characteristic polynomials of ; i.e., Also, . Now, the Chinese remainder theorem applied to the polynomial ring gives a polynomial satisfying the conditions
The condition, when spelled out, means that for some polynomial. Since is the zero map on, and agree on each ; i.e.,. Also then with. The condition ensures that and have no constant terms. This completes the proof of the algebraically closed field case.
If k is an arbitrary perfect field, let be the absolute Galois group of k. By the first part, we can choose polynomials over such that is the decomposition into the semisimple and nilpotent part. For each in,
Now, is a polynomial in ; so is. Thus, and commute. Also, the application of evidently preserves semisimplicity and nilpotency. Thus, by the uniqueness of decomposition, and. Hence, are -invariant; i.e., they are endomorphisms over k. Finally, since contains a -basis that spans the space containing, by the same argument, we also see that have coefficients in k. This completes the proof.
Short proof using abstract algebra
proves the existence of a decomposition as a consequence of the Wedderburn principal theorem.Let V be a finite-dimensional vector space over a perfect field k, an endomorphism and the subalgebra generated by x. Note that A is a commutative Artinian ring. The Wedderburn principal theorem states: for a finite-dimensional algebra A with the Jacobson radical J, if is separable, then the natural surjection splits; i.e., contains a semisimple subalgebra such that is an isomorphism. In the setup here, is separable since the base field is perfect and J is also the nilradical of A. There is then the vector-space decomposition. In particular, the endomorphism x can be written as where is in and in. Now, the image of x generates ; thus is semisimple and is a polynomial of x. Also, is nilpotent since is nilpotent and is a polynomial of x since is.
Nilpotency criterion
The Jordan decomposition can be used to characterize nilpotency of an endomorphism. Let k be an algebraically closed field of characteristic zero, the endomorphism ring of k over rational numbers and V a finite-dimensional vector space over k. Given an endomorphism, let be the Jordan decomposition. Then is diagonalizable; i.e., where each is the eigenspace for eigenvalue with multiplicity. Then for any let be the endomorphism such that is the multiplication by. Chevalley calls the replica of given by. Now,Proof: First, since is nilpotent,
If is the complex conjugation, this implies for every i. Otherwise, take to be a -linear functional followed by. Applying that to the above equation, one gets:
and, since are all real numbers, for every i. Varying the linear functionals then implies for every i.
A typical application of the above criterion is the proof of Cartan's criterion for solvability of a Lie algebra. It says: if is a Lie subalgebra over a field k of characteristic zero such that for each, then is solvable.
Proof: Without loss of generality, assume k is algebraically closed. By Lie's theorem and Engel's theorem, it suffices to show for each, is a nilpotent endomorphism of V. Write. Then we need to show:
is zero. Let. Note we have: and, since is the semisimple part of the Jordan decomposition of, it follows that is a polynomial without constant term in ; hence, and the same is true with in place of. That is,, which implies the claim given the assumption.
Counterexample to existence over an imperfect field
If the ground field is not perfect, then a Jordan–Chevalley decomposition may not exist. Example: Let p be a prime number, let k be imperfect of characteristic p, and choose a in k that is not a pth power. Let V = k/2, and let T be the k-linear operator given by multiplication by x on V. This has as its stable k-linear subspaces precisely the ideals of V viewed as a ring. Suppose T=S+N for commuting k-linear operators S and N that are respectively semisimple and nilpotent. Since S and N commute, they each commute with T=S+N and hence each acts k-linearly on V. Thus, each preserves the unique nonzero proper k-submodule J=V in V. But by semisimplicity of S, there would have to be an S-stable k-linear complement to J. However, by k-linearity, S and N are each given by multiplication against the respective polynomials s = S and n =N whose induced effects on the quotient V/ must be respectively x and 0 since this quotient is a field. Hence, s = x + h for some polynomial h, so it is easily seen that s generates V as a k-algebra and thus the S-stable k-linear subspaces of V are precisely the k-submodules. It follows that an S-stable complement to J is also a k-submodule of V, contradicting that J is the only nonzero proper k-submodule of V. Thus, there is no decomposition of T as a sum of commuting k-linear operators that are respectively semisimple and nilpotent.Analogous decompositions
The multiplicative version of the Jordan-Chevalley decomposition generalizes to a decomposition in a linear algebraic group, and the additive version of the decomposition generalizes to a decomposition in a Lie algebra.Lie algebras
Let denote the Lie algebra of the endomorphisms of a finite-dimensional vector space V over a perfect field. If is the Jordan decomposition, then is the Jordan decomposition of on the vector space. Indeed, first, and commute since. Second, in general, for each endomorphism, we have:- If, then, since is the difference of the left and right multiplications by y.
- If is semisimple, then is semisimple.
If is a finite-dimensional representation of a semisimple finite-dimensional complex Lie algebra, then preserves the Jordan decomposition in the sense: if, then and.
Real semisimple Lie algebras
In the formulation of Chevalley and Mostow, the additive decomposition states that an element X in a real semisimple Lie algebra g with Iwasawa decomposition g = k ⊕ a ⊕ n can be written as the sum of three commuting elements of the Lie algebra X = S + D + N, with S, D and N conjugate to elements in k, a and n respectively. In general the terms in the Iwasawa decomposition do not commute.Linear algebraic groups
Let be a linear algebraic group over a perfect field. Then, essentially by definition, there is a closed embedding. Now, to each element, by the multiplicative Jordan decomposition, there are a pair of a semisimple element and a unipotent element a priori in such that. But, as it turns out, the elements can be shown to be in and that they are independent of the embedding into ; i.e., the decomposition is intrinsic.When G is abelian, is then the direct product of the closed subgroup of the semisimple elements in G and that of unipotent elements.