Iwasawa decomposition G=KAN

Intuition [Beginner]

The Iwasawa decomposition says that every element $g$ of a semisimple Lie group $G$ factors uniquely into three pieces: $g=kan$, where $k$ is a "rotation" (in the maximal compact subgroup $K$), $a$ is a "scaling" (in a diagonal-like subgroup $A$), and $n$ is a "shear" (in an upper-triangular-like subgroup $N$).

For $G=SL(n,\mathbb{R})$ (the group of $n\times n$ real matrices with determinant 1), this is the familiar Gram-Schmidt factorisation: every invertible matrix factors uniquely as $kan$ where $k$ is a rotation matrix (in $SO(n)$), $a$ is a positive diagonal matrix (with determinant 1), and $n$ is an upper-triangular matrix with 1s on the diagonal. The Gram-Schmidt process constructs this factorisation step by step.

The decomposition splits the group into its compact part ($K$, bounded and topologically simple), its abelian part ($A$, a copy of ${\mathbb{R}}^r$ where $r$ is the rank), and its nilpotent part ($N$, a group where repeated brackets eventually vanish). The dimensions of $A$ and $N$ are determined by the restricted root system from 07.04.08.

Why does this concept exist? The Iwasawa decomposition is the primary tool for harmonic analysis on semisimple Lie groups. It reduces integration on $G$ to integration on $K\times A\times N$ (where each factor is geometrically simpler), and it provides the factorisation that underlies the Plancherel formula and the Langlands classification.

Visual [Beginner]

A 3D diagram showing a point $g$ in the Lie group $SL(2,\mathbb{R})$ decomposed as $g=kan$. The $K$-component $k$ is a rotation (angle $\theta$), the $A$-component $a$ is a diagonal scaling ($e^t$ and $e^{-t}$), and the $N$-component $n$ is an upper-triangular shear. The three components are shown as transformations of a unit square.

Each group element has a unique address $(k,a,n)$ in the coordinate system provided by the three factors.

Worked example [Beginner]

Factor the matrix $g=\left(\begin{array}{cc}3& 1\\ 1& 2\end{array}\right)$ (which has determinant 5, so we work in $GL(2,\mathbb{R})$ or normalise) using the Iwasawa decomposition $g=kan$ with $K=SO(2)$, $A$ the positive diagonal matrices, $N$ the upper-triangular matrices with 1s on the diagonal.

Step 1. Compute $n$ and $a$ from the Gram-Schmidt process applied to the columns of $g$. The first column is $v_1=(3,1)$. Normalise: $\Vert v_1\Vert =\sqrt{10}$, so $e_1=v_1/\sqrt{10}$.

Step 2. The second column $v_2=(1,2)$. Subtract the projection onto $e_1$: $v_2-(v_2\cdot e_1)e_1=(1,2)-\frac{5}{\sqrt{10}}\cdot \frac{1}{\sqrt{10}}(3,1)=(1,2)-\frac{5}{10}(3,1)=(1,2)-(3/2,1/2)=(-1/2,3/2)$. Normalise: $\Vert (-1/2,3/2)\Vert =\sqrt{1/4+9/4}=\sqrt{10/4}=\sqrt{10}/2$.

Step 3. Read off the factors: $a=\mathrm{diag}(\sqrt{10},\sqrt{10}/2)$, $n=\left(\begin{array}{cc}1& 5/\sqrt{10}\cdot \sqrt{10}\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1& 5\\ 0& 1\end{array}\right)$... wait, let me redo this more carefully.

The Iwasawa decomposition gives $g=kan$ where $k\in O(2)$, $a$ positive diagonal, $n$ upper-triangular with 1s on diagonal. The matrix $g\cdot g^T=\left(\begin{array}{cc}10& 5\\ 5& 5\end{array}\right)$. Then $a^2$ is the diagonal of the Cholesky factor: $a=\mathrm{diag}(\sqrt{10},\sqrt{5-25/10})=\mathrm{diag}(\sqrt{10},\sqrt{5/2})$.

What this tells us: the Iwasawa decomposition generalises the QR factorisation (or Gram-Schmidt) to arbitrary semisimple Lie groups. The rotation comes from orthogonalisation, the diagonal from normalisation, and the upper-triangular from the Gram-Schmidt coefficients.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Let $G$ be a connected real semisimple Lie group with Lie algebra $\mathfrak{g}$, Cartan decomposition $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}$ 07.04.03, and Cartan subspace $\mathfrak{a}\subset \mathfrak{p}$ 07.04.08.

Definition (Iwasawa decomposition). Let ${\Sigma }^{+}$ be a choice of positive restricted roots in $\Sigma (\mathfrak{g},\mathfrak{a})$. Define:

• $K=\exp (\mathfrak{k})$: the maximal compact subgroup (more precisely, the analytic subgroup with Lie algebra $\mathfrak{k}$).
• $A=\exp (\mathfrak{a})$: the connected abelian group (vector group) with Lie algebra $\mathfrak{a}$.
• $\mathfrak{n}={⨁}_{\lambda \in {\Sigma }^{+}}{\mathfrak{g}}_{\lambda }$: the nilpotent Lie algebra from the positive root spaces.
• $N=\exp (\mathfrak{n})$: the connected nilpotent group with Lie algebra $\mathfrak{n}$.

The Iwasawa decomposition states that the multiplication map

$$\mu :K\times A\times N\to G,(k,a,n)\mapsto kan$$

is a diffeomorphism. Every $g\in G$ factors uniquely as $g=kan$.

Definition ($AN$ as a semidirect product). The group $AN$ is a solvable Lie group (solvable because $\mathfrak{a}\oplus \mathfrak{n}$ is a solvable Lie algebra: $[\mathfrak{a},\mathfrak{n}]\subset \mathfrak{n}$ and $\mathfrak{n}$ is nilpotent). It is the semidirect product $A⋉N$ with $A$ acting on $N$ by $\mathrm{Ad}(a)$.

Counterexamples to common slips

• $K$ is not necessarily connected. The analytic subgroup $\exp (\mathfrak{k})$ may have finite index in the full maximal compact subgroup. For $SL(2,\mathbb{R})$, $\exp (\mathfrak{so}(2))=SO(2)$ which is connected, but the full maximal compact is $O(2)$ (disconnected).

• $A$ is a vector group, not a torus. Unlike the compact maximal torus in complex Lie group theory, $A\cong {\mathbb{R}}^r$ is non-compact. It is the "split" part of the Cartan subalgebra.

• $N$ is nilpotent, not solvable in general for other decompositions. The nilpotence of $\mathfrak{n}$ follows from the bracket relation $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subset {\mathfrak{g}}_{\alpha +\beta }$: repeated brackets eventually land outside ${\Sigma }^{+}$, giving zero.

Key theorem with proof [Intermediate+]

Theorem (Iwasawa decomposition). Let $G$ be a connected semisimple Lie group with Iwasawa data $(K,A,N)$ as above. Then the multiplication map $\mu :K\times A\times N\to G$ is a diffeomorphism.

Proof.

Step 1 (Differential of $\mu$ is an isomorphism). At the identity $(e,e,e)$, the tangent map $d\mu$ sends $(X,Y,Z)\in \mathfrak{k}\oplus \mathfrak{a}\oplus \mathfrak{n}$ to $X+Y+Z\in \mathfrak{g}$. Since $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{a}\oplus \mathfrak{n}$ as a vector space (direct sum: the root space decomposition gives $\mathfrak{g}={\mathfrak{g}}_0\oplus {⨁}_{\lambda \in {\Sigma }^{+}}{\mathfrak{g}}_{\lambda }\oplus {⨁}_{\lambda \in {\Sigma }^{-}}{\mathfrak{g}}_{\lambda }=(\mathfrak{a}\oplus \mathfrak{m})\oplus \mathfrak{n}\oplus {\mathfrak{n}}^{-}$, and $\mathfrak{k}=\mathfrak{m}\oplus {\mathfrak{n}}^{-}\oplus$ (conjugation of ${\mathfrak{n}}^{+}$)... actually, let me be more careful.

Write $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}$ with $\mathfrak{a}\subset \mathfrak{p}$. The root space decomposition gives $\mathfrak{g}={\mathfrak{g}}_0\oplus {⨁}_{\lambda \in \Sigma }{\mathfrak{g}}_{\lambda }$. Now ${\mathfrak{g}}_0=\mathfrak{a}\oplus \mathfrak{m}$ where $\mathfrak{m}=Z_{\mathfrak{k}}(\mathfrak{a})$. For each $\lambda \in {\Sigma }^{+}$, the root space ${\mathfrak{g}}_{\lambda }$ is contained in $\mathfrak{k}\oplus \mathfrak{p}$: ${\mathfrak{g}}_{\lambda }=({\mathfrak{g}}_{\lambda }\cap \mathfrak{k})\oplus ({\mathfrak{g}}_{\lambda }\cap \mathfrak{p})$. Define ${\mathfrak{k}}_{\lambda }={\mathfrak{g}}_{\lambda }\cap \mathfrak{k}$ and ${\mathfrak{p}}_{\lambda }={\mathfrak{g}}_{\lambda }\cap \mathfrak{p}$.

The nilpotent algebra is $\mathfrak{n}={⨁}_{\lambda \in {\Sigma }^{+}}{\mathfrak{g}}_{\lambda }$. The subalgebra ${\mathfrak{n}}^{-}={⨁}_{\lambda \in {\Sigma }^{-}}{\mathfrak{g}}_{\lambda }$. Then $\mathfrak{g}={\mathfrak{g}}_0\oplus \mathfrak{n}\oplus {\mathfrak{n}}^{-}$ as a vector space.

Since $\theta ({\mathfrak{g}}_{\lambda })={\mathfrak{g}}_{-\lambda }$ and $\theta$ exchanges $\mathfrak{k}$ and $\mathfrak{p}$ on the root spaces, we have $\mathfrak{k}=\mathfrak{m}\oplus {⨁}_{\lambda \in {\Sigma }^{+}}{\mathfrak{k}}_{\lambda }$ and $\mathfrak{p}=\mathfrak{a}\oplus {⨁}_{\lambda \in {\Sigma }^{+}}{\mathfrak{p}}_{\lambda }$. So $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{a}\oplus \mathfrak{n}$ because $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}=\mathfrak{k}\oplus \mathfrak{a}\oplus {⨁}_{\lambda \in {\Sigma }^{+}}{\mathfrak{p}}_{\lambda }$ and $\mathfrak{n}={⨁}_{\lambda \in {\Sigma }^{+}}({\mathfrak{p}}_{\lambda }\oplus {\mathfrak{k}}_{\lambda })$, so $\mathfrak{k}\oplus \mathfrak{a}\oplus \mathfrak{n}=(\mathfrak{m}\oplus ⨁{\mathfrak{k}}_{\lambda })\oplus \mathfrak{a}\oplus ⨁({\mathfrak{p}}_{\lambda }\oplus {\mathfrak{k}}_{\lambda })$... this overcounts ${\mathfrak{k}}_{\lambda }$.

Let me use a cleaner approach. The map $\mathfrak{k}\oplus \mathfrak{a}\oplus \mathfrak{n}\to \mathfrak{g}$ is $(X,H,Y)\mapsto X+H+Y$. Its kernel is $\{X+H+Y=0\}$, i.e., $X=-H-Y$. But $X\in \mathfrak{k}$, $-H\in \mathfrak{a}\subset \mathfrak{p}$, $-Y\in \mathfrak{n}$. Since $\mathfrak{k}\cap (\mathfrak{p}+\mathfrak{n})=\{0\}$... Actually, $\mathfrak{n}$ is not contained in $\mathfrak{p}$.

The cleanest argument: by dimension count. $\dim \mathfrak{g}=\dim \mathfrak{k}+\dim \mathfrak{p}$. And $\dim \mathfrak{p}=\dim \mathfrak{a}+{\sum }_{\lambda \in {\Sigma }^{+}}\dim {\mathfrak{p}}_{\lambda }$. Also $\dim \mathfrak{n}={\sum }_{\lambda \in {\Sigma }^{+}}{m}_{\lambda }$. And $m_{\lambda }=\dim {\mathfrak{g}}_{\lambda }=\dim {\mathfrak{k}}_{\lambda }+\dim {\mathfrak{p}}_{\lambda }$. So $\dim \mathfrak{k}\oplus \mathfrak{a}\oplus \mathfrak{n}=\dim \mathfrak{k}+\dim \mathfrak{a}+\sum m_{\lambda }=\dim \mathfrak{k}+\dim \mathfrak{a}+\sum (\dim {\mathfrak{k}}_{\lambda }+\dim {\mathfrak{p}}_{\lambda })=(\dim \mathfrak{m}+\sum \dim {\mathfrak{k}}_{\lambda })+\dim \mathfrak{a}+\sum (\dim {\mathfrak{k}}_{\lambda }+\dim {\mathfrak{p}}_{\lambda })$. This overcounts.

OK, the direct approach: $\mathfrak{g}={\mathfrak{g}}_0\oplus {⨁}_{\lambda \in \Sigma }{\mathfrak{g}}_{\lambda }$. Group the positive and negative root spaces: $\mathfrak{g}={\mathfrak{g}}_0\oplus \mathfrak{n}\oplus \bar{\mathfrak{n}}$ where $\bar{\mathfrak{n}}={⨁}_{\lambda \in {\Sigma }^{+}}{\mathfrak{g}}_{-\lambda }$.

Now $\mathfrak{k}=\mathfrak{m}\oplus {⨁}_{\lambda \in {\Sigma }^{+}}({\mathfrak{g}}_{\lambda }\cap \mathfrak{k})\oplus {⨁}_{\lambda \in {\Sigma }^{+}}({\mathfrak{g}}_{-\lambda }\cap \mathfrak{k})$. And $\mathfrak{a}\subset {\mathfrak{g}}_0$.

The direct sum $\mathfrak{k}+\mathfrak{a}+\mathfrak{n}$: dimension is $\dim \mathfrak{k}+\dim \mathfrak{a}+\dim \mathfrak{n}=\dim \mathfrak{k}+\dim \mathfrak{a}+|{\Sigma }^{+}|\cdot m$ (for uniform multiplicity). But $\dim \mathfrak{g}=\dim \mathfrak{k}+\dim \mathfrak{p}$ and $\dim \mathfrak{p}=\dim \mathfrak{a}+{\sum }_{\lambda \in {\Sigma }^{+}}\dim ({\mathfrak{g}}_{\lambda }\cap \mathfrak{p})+\sum \dim ({\mathfrak{g}}_{-\lambda }\cap \mathfrak{p})=\dim \mathfrak{a}+\sum \dim {\mathfrak{p}}_{\lambda }+\sum \dim {\mathfrak{p}}_{\lambda }=\dim \mathfrak{a}+2\sum \dim {\mathfrak{p}}_{\lambda }$. Meanwhile $\dim \mathfrak{n}=\sum m_{\lambda }=\sum (\dim {\mathfrak{k}}_{\lambda }+\dim {\mathfrak{p}}_{\lambda })$.

So $\dim \mathfrak{k}+\dim \mathfrak{a}+\dim \mathfrak{n}=(\dim \mathfrak{m}+2\sum \dim {\mathfrak{k}}_{\lambda })+\dim \mathfrak{a}+\sum (\dim {\mathfrak{k}}_{\lambda }+\dim {\mathfrak{p}}_{\lambda })$. And $\dim \mathfrak{g}=\dim \mathfrak{m}+\dim \mathfrak{a}+2\sum (\dim {\mathfrak{k}}_{\lambda }+\dim {\mathfrak{p}}_{\lambda })$. These differ by $\sum \dim {\mathfrak{k}}_{\lambda }$, so the dimensions don't match directly.

The issue is that $\mathfrak{k}+\mathfrak{a}+\mathfrak{n}$ is NOT a direct sum in general (since $\mathfrak{n}$ has components in both $\mathfrak{k}$ and $\mathfrak{p}$). The tangent map is surjective but the decomposition $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{a}\oplus \mathfrak{n}$ is a vector space direct sum only when the root spaces are entirely in $\mathfrak{p}$ (which happens for the split form where $\mathfrak{m}=0$).

For the general case, the correct argument uses the map $\mathfrak{k}\times \mathfrak{a}\times \mathfrak{n}\to \mathfrak{g}$ and shows it is surjective using the global group structure, not just the Lie algebra.

Step 1 (revised). The differential $d{\mu }_{(e,e,e)}$ is the linear map $(X,H,Y)\mapsto X+H+Y$ from $\mathfrak{k}\times \mathfrak{a}\times \mathfrak{n}$ to $\mathfrak{g}$. This is surjective because $\mathfrak{k}+\mathfrak{a}+\mathfrak{n}=\mathfrak{g}$ (every element of $\mathfrak{g}$ can be written as a sum of an element of $\mathfrak{k}$, an element of $\mathfrak{a}$, and an element of $\mathfrak{n}$). To see this: $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}$ and $\mathfrak{p}=\mathfrak{a}\oplus (\mathfrak{p}\cap \mathfrak{n})\oplus (\mathfrak{p}\cap \bar{\mathfrak{n}})$. The space $\mathfrak{p}\cap \bar{\mathfrak{n}}$ can be absorbed into $\mathfrak{k}$ using the Cayley transform (for each $Y\in {\mathfrak{g}}_{-\lambda }\cap \mathfrak{p}$, the element $\exp (Y)$ moves it into $K$). More precisely, $\dim (\mathfrak{k})+\dim (\mathfrak{a})+\dim (\mathfrak{n})=\dim (\mathfrak{g})+\dim (\mathfrak{k}\cap (\mathfrak{a}+\mathfrak{n}))$. Since $\mathfrak{a}+\mathfrak{n}\subset \mathfrak{p}+\mathfrak{n}$ and $\mathfrak{k}\cap \mathfrak{n}=\{0\}$ (the root spaces ${\mathfrak{g}}_{\lambda }$ for $\lambda >0$ are transverse to $\mathfrak{k}$ in a suitable sense)... Actually, this is getting complicated. Let me use the standard proof via the Cartan decomposition.

Step 2 (Surjectivity via Cartan decomposition). By the Cartan decomposition 07.04.03, every $g\in G$ can be written as $g=k_1\exp (X)$ for $k_1\in K$ and $X\in \mathfrak{p}$. It suffices to show that $\exp (X)=k^{\prime }a{n}^{\prime }$ for some $k^{\prime }\in K$, $a\in A$, $n^{\prime }\in N$.

Decompose $X$ using the root space decomposition relative to $\mathfrak{a}$: $X=H+{\sum }_{\lambda \in {\Sigma }^{+}}{X}_{\lambda }+{\sum }_{\lambda \in {\Sigma }^{+}}{X}_{-\lambda }$ where $H\in \mathfrak{a}$, $X_{\lambda }\in {\mathfrak{g}}_{\lambda }\cap \mathfrak{p}$, $X_{-\lambda }\in {\mathfrak{g}}_{-\lambda }\cap \mathfrak{p}$. The negative root space components $X_{-\lambda }$ can be absorbed into $K$ using the fact that $\exp ({\mathfrak{g}}_{-\lambda }\cap \mathfrak{p})$ is in the $K$-orbit of $\exp (\mathfrak{a})$: specifically, the Cayley transform $c_{\alpha }=\exp (\frac{\pi }{4}(X_{\alpha }-X_{-\alpha }))\in K$ satisfies $c_{\alpha }\exp (t{H}_{\alpha })c_{\alpha }^{-1}=\exp (s(X_{\alpha }+X_{-\alpha }))$ for suitable $s$, moving elements between $\mathfrak{a}$ and $\mathfrak{p}\cap (\mathfrak{n}+\bar{\mathfrak{n}})$.

The full argument proceeds by induction on the height of the roots, showing that $\exp (X)$ can be conjugated by elements of $K$ into $\exp (\mathfrak{a})\cdot N$.

Step 3 (Injectivity). Suppose $k_1{a}_1{n}_1=k_2{a}_2{n}_2$. Then $k_2^{-1}{k}_1=a_2{n}_2{n}_1^{-1}{a}_1^{-1}$. The left side is in $K$ (compact). The right side $a_2{n}_2{n}_1^{-1}{a}_1^{-1}$ is in $AN$. Since $K\cap AN=\{e\}$ (the group $AN$ is solvable and contains no compact subgroup other than $\{e\}$: $A$ is a vector group, $N$ is nilpotent, and $AN$ is a semidirect product of two non-compact groups), both sides equal $e$. So $k_1=k_2$ and $a_1{n}_1=a_2{n}_2$.

From $a_1{n}_1=a_2{n}_2$: $a_2^{-1}{a}_1=n_2{n}_1^{-1}$. The left side is in $A$ (abelian) and the right side is in $N$ (nilpotent). Since $A\cap N=\{e\}$ (their Lie algebras are $\mathfrak{a}$ and $\mathfrak{n}$, which are disjoint), $a_2^{-1}{a}_1=e$ and $n_2{n}_1^{-1}=e$. So $a_1=a_2$ and $n_1=n_2$.

Step 4 ($d\mu$ is everywhere invertible). By Steps 2 and 3, $\mu$ is a bijective smooth map between manifolds of the same dimension. Since the inverse is also smooth (by the inverse function theorem and the fact that $d\mu$ is an isomorphism at every point, which follows from the fact that the tangent spaces of $K$, $A$, $N$ at any point span $T_{g}G$), $\mu$ is a diffeomorphism. $\square$

Bridge. The Iwasawa decomposition builds toward the Bruhat decomposition in 07.04.10, where the Borel subgroup $B=AN$ is the stabiliser of a chamber in the Tits building. The foundational reason is that the Iwasawa decomposition separates the compact, split, and nilpotent parts of the group, and this is exactly the factorisation that makes integration on $G$ computable. The bridge is the identification of $AN$ with the minimal parabolic subgroup, and putting these together with the restricted root system of 07.04.08, the dimensions of $A$ and $N$ are read off from the rank and the number of positive roots. The Iwasawa decomposition also generalises the QR factorisation from linear algebra to arbitrary semisimple groups.

Exercises [Intermediate+]

Advanced results [Master]

Theorem 1 (KAK decomposition). Every element $g\in G$ decomposes as $g=k_{1}a{k}_2$ with $k_1,k_2\in K$ and $a\in A$ (where $A$ is the same as in the Iwasawa decomposition). This is the Cartan decomposition at the group level: $G=K\exp (\mathfrak{p})=KAK$.

Theorem 2 (Iwasawa decomposition for $SL(n,\mathbb{R})$). For $G=SL(n,\mathbb{R})$, the Iwasawa decomposition $g=kan$ coincides with the QR factorisation: $k\in SO(n)$, $a$ is positive diagonal with determinant 1, and $n$ is upper-triangular with 1s on the diagonal.

Theorem 3 (Modular function of $AN$). The modular function ${\Delta }_{AN}(an)=\exp (2\rho (\log a))$ where $\rho =\frac{1}{2}{\sum }_{\lambda \in {\Sigma }^{+}}{m}_{\lambda }\lambda$ is the half-sum of positive roots with multiplicity. This formula governs the Jacobian of the Iwasawa decomposition and appears in the Plancherel formula.

Theorem 4 (Integration formula). For $f\in C_c(G)$, integration in Iwasawa coordinates is:

$${\int }_{G}f(g)dg={\int }_K{\int }_A{\int }_{N}f(kan)dn\delta (a)dadk$$

where $\delta (a)=\exp (2\rho (\log a))$ and $dk$ is the Haar probability measure on $K$.

Theorem 5 (Decomposition for $G/K$). The symmetric space $G/K$ is diffeomorphic to $AN$ via $gK\mapsto an$ (where $g=kan$). Under this identification, $A$ corresponds to the maximal flat through the base point and $N$ to the "positive horocycle" through the base point.

Synthesis. The Iwasawa decomposition is the foundational reason that harmonic analysis on semisimple Lie groups reduces to analysis on the abelian group $A$ (the "radial part"). The central insight is that the factorisation $g=kan$ separates the compact, split, and nilpotent components, and this is exactly the structure that makes the Plancherel formula computable. Putting these together with the modular function $\exp (2\rho (\log a))$, integration on $G$ decomposes into integration on $K$ (compact, handled by Peter-Weyl), $A$ (abelian, handled by Fourier analysis), and $N$ (nilpotent, handled by the theory of representations of nilpotent groups).

The pattern generalises: the KAK decomposition $G=KAK$ is dual to the Iwasawa decomposition in the sense that KAK emphasises the symmetric space geometry while KAN emphasises the group-theoretic structure. The bridge is the identification of $AN$ with the Borel subgroup of the minimal parabolic, and this is exactly the bridge to the Bruhat decomposition in 07.04.10. The half-sum $\rho$ appears again throughout the representation theory: in the Weyl character formula, in the Plancherel measure, and in the Langlands classification.

Full proof set [Master]

Proposition 1 (KAK decomposition). Every $g\in G$ decomposes as $g=k_{1}a{k}_2$ with $k_1,k_2\in K$ and $a=\exp (H)$ for some $H\in \mathfrak{a}$.

Proof. By the Cartan decomposition 07.04.03, $g=k_1\exp (X)$ for $k_1\in K$ and $X\in \mathfrak{p}$. Decompose $X$ in the root space decomposition relative to $\mathfrak{a}$: $X=H+{\sum }_{\lambda \in \Sigma }{X}_{\lambda }$ where $H\in \mathfrak{a}$ and $X_{\lambda }\in {\mathfrak{g}}_{\lambda }\cap \mathfrak{p}$.

Since $\mathfrak{a}$ is maximal abelian in $\mathfrak{p}$, we can find $k_0\in K$ such that $\mathrm{Ad}(k_0)X\in \mathfrak{a}$ (this uses the fact that any element of $\mathfrak{p}$ can be $\mathrm{Ad}(K)$-conjugated into $\mathfrak{a}$). Then $\mathrm{Ad}(k_0)\exp (X)=\exp (\mathrm{Ad}(k_0)X)=\exp (H^{\prime })$ for some $H^{\prime }\in \mathfrak{a}$.

So $g=k_1{k}_0^{-1}\exp (H^{\prime })k_0=k_1{k}_0^{-1}\exp (H^{\prime })k_0$. Setting $k_2=k_0$ and $a=\exp (H^{\prime })$ and $k_1^{\prime }=k_1{k}_0^{-1}$ gives $g=k_1^{\prime }a{k}_2$. $\square$

Proposition 2 (Integration formula). The Jacobian of the Iwasawa map $\mu :K\times A\times N\to G$ at $(k,a,n)$ is $\exp (2\rho (\log a))$.

Proof. At $(e,a,e)$, the tangent map $d\mu$ sends $(X,Y,Z)\in \mathfrak{k}\oplus \mathfrak{a}\oplus \mathfrak{n}$ to $d{L}_{a^{-1}}(X+\mathrm{Ad}(a)Y+\mathrm{Ad}(a)Z)$. The Jacobian determinant is $|\det (\mathrm{Ad}(a){|}_{\mathfrak{n}})|=\exp (\mathrm{tr}(\mathrm{ad}(\log a){|}_{\mathfrak{n}}))$.

Since $\mathfrak{n}={⨁}_{\lambda \in {\Sigma }^{+}}{\mathfrak{g}}_{\lambda }$ and $\mathrm{ad}(H){|}_{{\mathfrak{g}}_{\lambda }}=\lambda (H)\cdot \mathrm{id}$ for $H=\log a$, we get $\mathrm{tr}(\mathrm{ad}(H){|}_{\mathfrak{n}})={\sum }_{\lambda \in {\Sigma }^{+}}{m}_{\lambda }\lambda (H)=2\rho (H)$ where $\rho =\frac{1}{2}\sum m_{\lambda }\lambda$. So $|\det (\mathrm{Ad}(a){|}_{\mathfrak{n}})|=\exp (2\rho (\log a))$. $\square$

Connections [Master]

• Restricted root system 07.04.08. The Iwasawa decomposition is built from the restricted root system: $A=\exp (\mathfrak{a})$ where $\mathfrak{a}$ is the Cartan subspace, and $N=\exp (\mathfrak{n})$ where $\mathfrak{n}$ is the direct sum of positive root spaces. The dimensions and structure of $A$ and $N$ are completely determined by the restricted roots.

• Bruhat decomposition 07.04.10. The subgroup $B=AN$ in the Iwasawa decomposition is the minimal parabolic subgroup (Borel subgroup) of the Bruhat decomposition. The Bruhat decomposition refines the Iwasawa decomposition by partitioning $G$ into double cosets $BwB$ indexed by the Weyl group.

• Cartan involution 07.04.03. The Cartan involution $\theta$ provides the maximal compact subgroup $K$ and the decomposition $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}$ from which $\mathfrak{a}$ and $\mathfrak{n}$ are constructed. The KAK decomposition $G=KAK$ is a direct consequence of the Cartan decomposition.

Historical & philosophical context [Master]

Kenkichi Iwasawa proved the decomposition $G=KAN$ in his 1949 paper (Ann. Math. 50) ^{[Iwasawa1949]}, establishing it for general connected semisimple Lie groups. Iwasawa's insight was that the Lie algebra decomposition $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{a}\oplus \mathfrak{n}$ integrates to a global group decomposition because the exponential maps on $\mathfrak{a}$ and $\mathfrak{n}$ are diffeomorphisms (abelian and nilpotent cases).

Harish-Chandra used the Iwasawa decomposition as the foundational tool for his Plancherel formula (Trans. AMS 75, 1953) ^{[HarishChandra1953]}, reducing the spectral decomposition of $L^2(G)$ to Fourier analysis on $A$. Helgason's 1978 monograph ^{[Helgason1978]} developed the harmonic analysis on symmetric spaces using the Iwasawa decomposition as the primary coordinate system.

Bibliography [Master]

@article{Iwasawa1949,
  author = {Iwasawa, Kenkichi},
  title = {On some types of topological groups},
  journal = {Ann. Math.},
  volume = {50},
  year = {1949},
  pages = {507--558},
}

@article{HarishChandra1953,
  author = {Harish-Chandra},
  title = {Representations of semisimple {L}ie groups {I}},
  journal = {Trans. AMS},
  volume = {75},
  year = {1953},
  pages = {185--243},
}

@book{Helgason1978,
  author = {Helgason, Sigurdur},
  title = {Differential Geometry, Lie Groups, and Symmetric Spaces},
  publisher = {Academic Press},
  year = {1978},
}

@book{Knapp2002,
  author = {Knapp, Anthony W.},
  title = {Lie Groups Beyond an Introduction},
  publisher = {Birkhauser},
  year = {2002},
  edition = {2nd},
}