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 important in the study of algebraic groups. 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.
Decomposition of endomorphisms
Consider linear operators on a finite-dimensional vector space over a perfect field. An operator T is semisimple if every T-invariant subspace has a complementary T-invariant subspace (if the underlying field is algebraically closed, this is the same as the requirement that the operator be diagonalizable). 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:
- x = xss + xn,
where xss is semisimple, xn is nilpotent, and xss and xn commute. If such a decomposition exists it is unique, and xss and xn are in fact expressible as polynomials in x, (Humphreys 1972, Prop. 4.2, p. 17).
If x is an invertible operator, then a multiplicative Jordan–Chevalley decomposition expresses x as a product:
- x = xss · xu,
where xss is semisimple, xu is unipotent, and xss and xu commute. Again, if such a decomposition exists it is unique, and xss and xu are expressible as polynomials in x.
For endomorphisms of a finite dimensional vector space whose characteristic polynomial splits into linear factors over the ground field (which always happens if that is an algebraically closed field), the Jordan–Chevalley decomposition exists and has a simple description in terms of the Jordan normal form. If x is in the Jordan normal form, then xss is the endomorphism whose matrix on the same basis contains just the diagonal terms of x, and xn is the endomorphism whose matrix on that basis 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.
Decomposition in a real semisimple Lie algebra
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.
Decomposition in a real semisimple Lie group
The multiplicative decomposition states that if g is an element of the corresponding connected semisimple Lie group G with corresponding Iwasawa decomposition G = KAN, then g can be written as the product of three commuting elements g = sdu with s, d and u conjugate to elements of K, A and N respectively. In general the terms in the Iwasawa decomposition g = kan do not commute.
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[x]/(xp-a)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 (just over k, which is weaker than semisimplicity over an algebraic closure of k) and nilpotent. Since S and N commute, they each commute with T=S+N and hence each acts k[x]-linearly on V. Thus, each preserves the unique nonzero proper k[x]-submodule J=(xp-a)V in V. But by semisimplicity of S, there would have to be an S-stable k-linear complement to J. However, by k[x]-linearity, S and N are each given by multiplication against the respective polynomials s = S(1) and n =N(1) whose induced effects on the quotient V/(xp-a) must be respectively x and 0 since this quotient is a field. Hence, s = x + (xp-a)h(x) for some polynomial h(x) (which only matters modulo (xp-a)), so it is easily seen[dubious ] that s generates V as a k-algebra and thus the S-stable k-linear subspaces of V are precisely the k[x]-submodules. It follows that an S-stable complement to J is also a k[x]-submodule of V, contradicting that J is the only nonzero proper k[x]-submodule of V. Thus, there is no decomposition of T as a sum of commuting k-linear operators that are respectively semisimple and nilpotent.
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