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 (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 x^m 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 = x_s + x_n,

where x_s is semisimple, x_n is nilpotent, and x_s and x_n commute. Over a perfect field,[1] such a decomposition exists (cf. #Proof of uniqueness and existence), the decomposition is unique, and the x_s and x_n are polynomials in x with no constant terms.[2][3] In particular, for any such decomposition over a perfect field, an operator that commutes with x also commutes with x_s and x_n.

If x is an invertible operator, then a multiplicative Jordan–Chevalley decomposition expresses x as a product

x = x_s · x_u,

where x_s is semisimple, x_u is unipotent, and x_s and x_u commute. Again, over a perfect field, such a decomposition exists, the decomposition is unique, and x_s and x_u are polynomials in x. The multiplicative version of the decomposition follows from the additive one since, as ${\displaystyle x_{s}}$ is easily seen to be invertible,

${\displaystyle x=x_{s}+x_{n}=x_{s}(1+x_{s}^{-1}x_{n})}$

and ${\displaystyle 1+x_{s}^{-1}x_{n}}$ is unipotent. (Conversely, by the same type of argument, one can deduce the additive version from the multiplicative one.)

If x is written in Jordan normal form (with respect to some basis) then x_s is the endomorphism whose matrix contains just the diagonal terms of x, and x_n is the endomorphism whose matrix contains just the off-diagonal terms; x_u 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 ${\displaystyle x_{s},x_{n}}$ are polynomial in x: if ${\displaystyle x=x_{s}'+x_{n}'}$ is another decomposition such that ${\displaystyle x'_{s}}$ and ${\displaystyle x'_{n}}$ commute, then ${\displaystyle x_{s}-x_{s}'=x_{n}'-x_{n}}$, and both ${\displaystyle x_{s}',x_{n}'}$ commute with x, hence with ${\displaystyle x_{s},x_{n}}$. Now, the sum of commuting semisimple (resp. nilpotent) endomorphisms is again semisimple (resp. nilpotent). Since the only operator which is both semisimple and nilpotent is the zero operator it follows that ${\displaystyle x_{s}=x_{s}'}$ and ${\displaystyle x_{n}=x_{n}'}$.

We show the existence. Let V be a finite-dimensional vector space over a perfect field k and ${\displaystyle x:V\to V}$ an endomorphism.

First assume the base field k is algebraically closed. Then the vector space V has the direct sum decomposition ${\displaystyle V=\bigoplus _{i=1}^{r}V_{i}}$ where each ${\displaystyle V_{i}}$ is the kernel of ${\displaystyle (x-\lambda _{i}I)^{m_{i}}}$, the generalized eigenspace and x stabilizes ${\displaystyle V_{i}}$, meaning ${\displaystyle x\cdot V_{i}\subset V_{i}}$. Now, define ${\displaystyle x_{s}:V\to V}$ so that, on each ${\displaystyle V_{i}}$, it is the scalar multiplication by ${\displaystyle \lambda _{i}}$. Note that, in terms of a basis respecting the direct sum decomposition, ${\displaystyle x_{s}}$ is a diagonal matrix; hence, it is a semisimple endomorphism. Since ${\displaystyle x-x_{s}:V_{i}\to V_{i}}$ is then ${\displaystyle x-\lambda _{i}I:V_{i}\to V_{i}}$ whose ${\displaystyle m_{i}}$-th power is zero, we also have that ${\displaystyle x_{n}:=x-x_{s}}$ is nilpotent, establishing the existence of the decomposition.

(Choosing a basis carefully on each ${\displaystyle V_{i}}$, one can then put x in the Jordan normal form and ${\displaystyle x_{s},x_{n}}$ are the diagonal and the off-diagonal parts of the normal form. But this is not needed here.)

The fact that ${\displaystyle x_{s},x_{n}}$ are polynomials in x follows from the Chinese remainder theorem. Indeed, let ${\displaystyle f(t)=\operatorname {det} (tI-x)}$ be the characteristic polynomial of x. Then it is the product of the characteristic polynomials of ${\displaystyle x:V_{i}\to V_{i}}$; i.e., ${\displaystyle f(t)=\prod _{i=1}^{r}(t-\lambda _{i})^{d_{i}},d_{i}=\dim V_{i}.}$ Also, ${\displaystyle d_{i}\geq m_{i}}$ (because, in general, a nilpotent matrix is killed when raised to the size of the matrix). Now, the Chinese remainder theorem applied to the polynomial ring ${\displaystyle k[t]}$ gives a polynomial ${\displaystyle p(t)}$ satisfying the conditions

${\displaystyle p(t)\equiv 0{\bmod {t}},\,p(t)\equiv \lambda _{i}{\bmod {(}}t-\lambda _{i})^{d_{i}}}$ (for all i).

(There is a redundancy in the conditions if some ${\displaystyle \lambda _{i}}$ is zero but that is not an issue; just remove it from the conditions.)

The condition ${\displaystyle p(t)\equiv \lambda _{i}{\bmod {(}}t-\lambda _{i})^{d_{i}}}$, when spelled out, means that ${\displaystyle p(t)-\lambda _{i}=g_{i}(t)(t-\lambda _{i})^{d_{i}}}$ for some polynomial ${\displaystyle g_{i}(t)}$. Since ${\displaystyle (x-\lambda _{i}I)^{d_{i}}}$ is the zero map on ${\displaystyle V_{i}}$, ${\displaystyle p(x)}$ and ${\displaystyle x_{s}}$ agree on each ${\displaystyle V_{i}}$; i.e., ${\displaystyle p(x)=x_{s}}$. Also then ${\displaystyle q(x)=x_{n}}$ with ${\displaystyle q(t)=t-p(t)}$. The condition ${\displaystyle p(t)\equiv 0{\bmod {t}}}$ ensures that ${\displaystyle p(t)}$ and ${\displaystyle q(t)}$ have no constant terms. This completes the proof of the algebraically closed field case.

If k is an arbitrary perfect field, let ${\displaystyle \Gamma =\operatorname {Gal} ({\overline {k}}/k)}$ be the absolute Galois group of k. By the first part, we can choose polynomials ${\displaystyle p,q}$ over ${\displaystyle {\overline {k}}}$ such that ${\displaystyle x=p(x)+q(x)}$ is the decomposition into the semisimple and nilpotent part. For each ${\displaystyle \sigma }$ in ${\displaystyle \Gamma }$,

${\displaystyle x=\sigma (x)=\sigma (p(x))+\sigma (q(x))=p(x)+q(x).}$

Now, ${\displaystyle \sigma (p(x))=\sigma (p)(x)}$ is a polynomial in ${\displaystyle x}$; so is ${\displaystyle \sigma (q(x))}$. Thus, ${\displaystyle \sigma (p(x))}$ and ${\displaystyle \sigma (q(x))}$ commute. Also, the application of ${\displaystyle \sigma }$ evidently preserves semisimplicity and nilpotency. Thus, by the uniqueness of decomposition (over ${\displaystyle {\overline {k}}}$), ${\displaystyle \sigma (p(x))=p(x)}$ and ${\displaystyle \sigma (q(x))=q(x)}$. Hence, ${\displaystyle x_{s}=p(x),x_{n}=q(x)}$ are ${\displaystyle \Gamma }$-invariant; i.e., they are endomorphisms (represented by matrices) over k. Finally, since ${\displaystyle \{1,x,x^{2},\dots \}}$ contains a ${\displaystyle {\overline {k}}}$-basis that spans the space containing ${\displaystyle x_{s},x_{n}}$, by the same argument, we also see that ${\displaystyle p,q}$ have coefficients in k. This completes the proof. ${\displaystyle \square }$

Short proof using abstract algebra

(Jacobson 1979) harv error: no target: CITEREFJacobson1979 (help) proves the existence of a decomposition as a consequence of the Wedderburn principal theorem. (This approach is not only short but also makes the role of the assumption that the base field be perfect clearer.)

Let V be a finite-dimensional vector space over a perfect field k, ${\displaystyle x:V\to V}$ an endomorphism and ${\displaystyle A=k[x]\subset \operatorname {End} (V)}$ 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 ${\displaystyle A/J}$ is separable, then the natural surjection ${\displaystyle p:A\to A/J}$ splits; i.e., ${\displaystyle A}$ contains a semisimple subalgebra ${\displaystyle B}$ such that ${\displaystyle p|_{B}:B{\overset {\sim }{\to }}A/J}$ is an isomorphism.[4] In the setup here, ${\displaystyle A/J}$ is separable since the base field is perfect (so the theorem is applicable) and J is also the nilradical of A. There is then the vector-space decomposition ${\displaystyle A=B\oplus J}$. In particular, the endomorphism x can be written as ${\displaystyle x=x_{s}+x_{n}}$ where ${\displaystyle x_{s}}$ is in ${\displaystyle B}$ and ${\displaystyle x_{n}}$ in ${\displaystyle J}$. Now, the image of x generates ${\displaystyle A/J\simeq B}$; thus ${\displaystyle x_{s}}$ is semisimple and is a polynomial of x. Also, ${\displaystyle x_{n}}$ is nilpotent since ${\displaystyle J}$ is nilpotent and is a polynomial of x since ${\displaystyle x_{s}}$ is. ${\displaystyle \square }$

Nilpotency criterion

The Jordan decomposition can be used to characterize nilpotency of an endomorphism. Let k be an algebraically closed field of characteristic zero, ${\displaystyle E=\operatorname {End} _{\mathbb {Q} }(k)}$ the endomorphism ring of k over rational numbers and V a finite-dimensional vector space over k. Given an endomorphism ${\displaystyle x:V\to V}$, let ${\displaystyle x=s+n}$ be the Jordan decomposition. Then ${\displaystyle s}$ is diagonalizable; i.e., ${\displaystyle V=\bigoplus V_{i}}$ where each ${\displaystyle V_{i}}$ is the eigenspace for eigenvalue ${\displaystyle \lambda _{i}}$ with multiplicity ${\displaystyle m_{i}}$. Then for any ${\displaystyle \varphi \in E}$ let ${\displaystyle \varphi (s):V\to V}$ be the endomorphism such that ${\displaystyle \varphi (s):V_{i}\to V_{i}}$ is the multiplication by ${\displaystyle \varphi (\lambda _{i})}$. Chevalley calls ${\displaystyle \varphi (s)}$ the replica of ${\displaystyle s}$ given by ${\displaystyle \varphi }$. (For example, if ${\displaystyle k=\mathbb {C} }$, then the complex conjugate of an endomorphism is an example of a replica.) Now,

Nilpotency criterion — [5]${\displaystyle x}$ is nilpotent (i.e., ${\displaystyle s=0}$) if and only if ${\displaystyle \operatorname {tr} (x\varphi (s))=0}$ for every ${\displaystyle \varphi \in E}$. Also, if ${\displaystyle k=\mathbb {C} }$, then it suffices the condition holds for ${\displaystyle \varphi =}$ complex conjugation.

Proof: First, since ${\displaystyle n\varphi (s)}$ is nilpotent,

${\displaystyle 0=\operatorname {tr} (x\varphi (s))=\sum _{i}\operatorname {tr} (s\varphi (s)|V_{i})=\sum _{i}m_{i}\lambda _{i}\varphi (\lambda _{i})}$.

If ${\displaystyle \varphi }$ is the complex conjugation, this implies ${\displaystyle \lambda _{i}=0}$ for every i. Otherwise, take ${\displaystyle \varphi }$ to be a ${\displaystyle \mathbb {Q} }$-linear functional ${\displaystyle \varphi :k\to \mathbb {Q} }$ followed by ${\displaystyle \mathbb {Q} \hookrightarrow k}$. Applying that to the above equation, one gets:

${\displaystyle \sum _{i}m_{i}\varphi (\lambda _{i})^{2}=0}$

and, since ${\displaystyle \varphi (\lambda _{i})}$ are all real numbers, ${\displaystyle \varphi (\lambda _{i})=0}$ for every i. Varying the linear functionals then implies ${\displaystyle \lambda _{i}=0}$ for every i. ${\displaystyle \square }$

A typical application of the above criterion is the proof of Cartan's criterion for solvability of a Lie algebra. It says: if ${\displaystyle {\mathfrak {g}}\subset {\mathfrak {gl}}(V)}$ is a Lie subalgebra over a field k of characteristic zero such that ${\displaystyle \operatorname {tr} (xy)=0}$ for each ${\displaystyle x\in {\mathfrak {g}},y\in D{\mathfrak {g}}=[{\mathfrak {g}},{\mathfrak {g}}]}$, then ${\displaystyle {\mathfrak {g}}}$ is solvable.

Proof:[6] Without loss of generality, assume k is algebraically closed. By Lie's theorem and Engel's theorem, it suffices to show for each ${\displaystyle x\in D{\mathfrak {g}}}$, ${\displaystyle x}$ is a nilpotent endomorphism of V. Write ${\displaystyle x=\sum _{i}[x_{i},y_{i}]}$. Then we need to show:

${\displaystyle \operatorname {tr} (x\varphi (s))=\sum _{i}\operatorname {tr} ([x_{i},y_{i}]\varphi (s))=\sum _{i}\operatorname {tr} (x_{i}[y_{i},\varphi (s)])}$

is zero. Let ${\displaystyle {\mathfrak {g}}'={\mathfrak {gl}}(V)}$. Note we have: ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(x):{\mathfrak {g}}\to D{\mathfrak {g}}}$ and, since ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(s)}$ is the semisimple part of the Jordan decomposition of ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(x)}$, it follows that ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(s)}$ is a polynomial without constant term in ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(x)}$; hence, ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(s):{\mathfrak {g}}\to D{\mathfrak {g}}}$ and the same is true with ${\displaystyle \varphi (s)}$ in place of ${\displaystyle s}$. That is, ${\displaystyle [\varphi (s),{\mathfrak {g}}]\subset D{\mathfrak {g}}}$, which implies the claim given the assumption. ${\displaystyle \square }$

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[x]/(x^p-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=(x^p-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/(x^p-a) must be respectively x and 0 since this quotient is a field. Hence, s = x + (x^p-a)h(x) for some polynomial h(x) (which only matters modulo (x^p-a)), 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[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.

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 ${\displaystyle {\mathfrak {gl}}(V)}$ denote the Lie algebra of the endomorphisms of a finite-dimensional vector space V over a perfect field. If ${\displaystyle x=x_{s}+x_{n}}$ is the Jordan decomposition, then ${\displaystyle \operatorname {ad} (x)=\operatorname {ad} (x_{s})+\operatorname {ad} (x_{n})}$ is the Jordan decomposition of ${\displaystyle \operatorname {ad} (x)}$ on the vector space ${\displaystyle {\mathfrak {gl}}(V)}$. Indeed, first, ${\displaystyle \operatorname {ad} (x_{s})}$ and ${\displaystyle \operatorname {ad} (x_{n})}$ commute since ${\displaystyle [\operatorname {ad} (x_{s}),\operatorname {ad} (x_{n})]=\operatorname {ad} ([x_{s},x_{n}])=0}$. Second, in general, for each endomorphism ${\displaystyle y\in {\mathfrak {gl}}(V)}$, we have:

1. If ${\displaystyle y^{m}=0}$, then ${\displaystyle \operatorname {ad} (y)^{2m-1}=0}$, since ${\displaystyle \operatorname {ad} (y)}$ is the difference of the left and right multiplications by y.
2. If ${\displaystyle y}$ is semisimple, then ${\displaystyle \operatorname {ad} (y)}$ is semisimple.[7]

Hence, by uniqueness, ${\displaystyle \operatorname {ad} (x)_{s}=\operatorname {ad} (x_{s})}$ and ${\displaystyle \operatorname {ad} (x)_{n}=\operatorname {ad} (x_{n})}$.

If ${\displaystyle \pi$ :{\mathfrak {g}}\to {\mathfrak {gl}}(V)} is a finite-dimensional representation of a semisimple finite-dimensional complex Lie algebra, then ${\displaystyle \pi }$ preserves the Jordan decomposition in the sense: if ${\displaystyle x=x_{s}+x_{n}}$, then ${\displaystyle \pi (x_{s})=\pi (x)_{s}}$ and ${\displaystyle \pi (x_{n})=\pi (x)_{n}}$.[8]

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 ${\displaystyle G}$ be a linear algebraic group over a perfect field. Then, essentially by definition, there is a closed embedding ${\displaystyle G\hookrightarrow \mathbf {GL} _{n}}$. Now, to each element ${\displaystyle g\in G}$, by the multiplicative Jordan decomposition, there are a pair of a semisimple element ${\displaystyle g_{s}}$ and a unipotent element ${\displaystyle g_{u}}$ a priori in ${\displaystyle \mathbf {GL} _{n}}$ such that ${\displaystyle g=g_{s}g_{u}=g_{u}g_{s}}$. But, as it turns out,[9] the elements ${\displaystyle g_{s},g_{u}}$ can be shown to be in ${\displaystyle G}$ (i.e., they satisfy the defining equations of G) and that they are independent of the embedding into ${\displaystyle \mathbf {GL} _{n}}$; i.e., the decomposition is intrinsic.

When G is abelian, ${\displaystyle G}$ is then the direct product of the closed subgroup of the semisimple elements in G and that of unipotent elements.[10]

Real semisimple Lie groups

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.

References

1. In fact, the proof goes through if the quotient ${\displaystyle k[x]/{\text{nil}}}$ is a separable algebra; see #Short proof using abstract algebra.
2. Humphreys 1972, Prop. 4.2, p. 17 for the algebraically closed field case.
3. Waterhouse, Ch. 9, Exercise 1. harvnb error: no target: CITEREFWaterhouse (help)
4. https://books.google.com/books?id=ZKGq4IQHhHUC&pg=PA143&dq=wedderburn+principal+theorem&hl=en&sa=X&redir_esc=y#v=onepage&q=wedderburn%20principal%20theorem&f=false
5. Serre, LA 5.17. Lemma 6.7. harvnb error: no target: CITEREFSerre (help) The endomorphism
6. Serre, LA 5.19. Theorem 7.1. harvnb error: no target: CITEREFSerre (help)
7. This is not easy to see but is shown in the proof of (Jacobson, Ch. III, § 7, Theorem 11.) harv error: no target: CITEREFJacobson (help). Editorial note: we need to add a discussion of this matter to “semisimple operator”.
8. Fulton & Harris, Theorem 9.20. harvnb error: no target: CITEREFFultonHarris (help)
9. Waterhouse, Theorem 9.2. harvnb error: no target: CITEREFWaterhouse (help)
10. Waterhouse, Theorem 9.3. harvnb error: no target: CITEREFWaterhouse (help)

• Chevalley, Claude (1951), Théorie des groupes de Lie. Tome II. Groupes algébriques, Hermann
• Fulton, William; Harris, Joe (1991). Representation theory. A first course. Graduate Texts in Mathematics, Readings in Mathematics. 129. New York: Springer-Verlag. doi:10.1007/978-1-4612-0979-9. ISBN 978-0-387-97495-8. MR 1153249. OCLC 246650103.
• Helgason, Sigurdur (1978), Differential geometry, Lie groups, and symmetric spaces, Academic Press, ISBN 0-8218-2848-7
• Humphreys, James E. (1981), Linear Algebraic Groups, Graduate texts in mathematics, 21, Springer, ISBN 0-387-90108-6
• Humphreys, James E. (1972), Introduction to Lie Algebras and Representation Theory, Springer, ISBN 978-0-387-90053-7
• Jacobson, Nathan, Lie algebras, Republication of the 1962 original. Dover Publications, Inc., New York, 1979. ISBN 0-486-63832-4
• Lazard, M. (1954), "Théorie des répliques. Critère de Cartan (Exposé No. 6)", Séminaire "Sophus Lie", 1
• Mostow, G. D. (1954), "Factor spaces of solvable groups", Ann. of Math., 60: 1–27, doi:10.2307/1969700
• Mostow, G. D. (1973), Strong rigidity of locally symmetric spaces, Annals of Mathematics Studies, 78, Princeton University Press
• Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, MR 1878556, Zbl 0984.00001
• Serre, Jean-Pierre (1992), Lie algebras and Lie groups: 1964 lectures given at Harvard University, Lecture Notes in Mathematics, 1500 (2nd ed.), Springer-Verlag, ISBN 978-3-540-55008-2
• Varadarajan, V. S. (1984), Lie groups, Lie algebras, and their representations, Graduate Texts in Mathematics, 102, Springer-Verlag, ISBN 0-387-90969-9
• Waterhouse, William (1979), Introduction to affine group schemes, Graduate Texts in Mathematics, 66, Berlin, New York: Springer-Verlag, doi:10.1007/978-1-4612-6217-6, ISBN 978-0-387-90421-4, MR 0547117