I am trying to find a certain proof of polar decomposition of complex matrices which I think should exist more generally for a certain class of Lie groups. Recall that the polar decomposition of a nonsingular complex matrix $A \in \text{GL}_n(\mathbb{C})$ is an expression $A = UP$, where $U \in \text{GL}_n(\mathbb{C})$ is unitary and $P \in \text{GL}_n(\mathbb{C})$ is positive definite.

My attempt arises with a well known analogy between four classes of matrices and four sets of complex numbers:

\begin{array}{|c|c|} \hline \text{Subset of }\text{Mat}_{n \times n } (\mathbb{C}) & \text{Corresponding set in } \mathbb{C} \text{ } \\ \hline \text{Hermitian matrices} & \text{Real numbers}\\ \hline \text{Skew-Hermitian matrices} & \text{Imaginary numbers} \\ \hline \text{Hermitian positive definite matrices} & \text{Positive real numbers} \\ \hline \text{Unitary matrices} & \text{The unit circle in } \mathbb{C}\\ \hline \end{array}

The analogy is justified through the observation that each set of matrices in the left column is precisely the set of unitarily diagonalizable matrices with eigenvalues in the corresponding set on the right. Moreover, the exponential map of matrices sends hermitian matrices to positive definite matrices and skew-hermitian to unitary matrices, a further similarity.

In showing polar decomposition, we take from the case for $\mathbb{C}$; $\text{exp} : \mathbb{C} \rightarrow \mathbb{C}^*$ sends the reals to the positive reals, and imaginary numbers to the unit circle. Taking $\text{exp}$ of $z = \text{Re}(z) + i \ \text{Im}(z)$ gives us the polar decomposition $\text{exp}(z) = e^{\text{Re}(z)} e^{i\ \text{Im}(z)}$ into a positive real number and a complex number on the unit circle. This analogy is presumably where polar decomposition gets its name.

Put $\mathfrak{g} = \text{Mat}_{n \times n } (\mathbb{C})$, the Lie algebra of $G = \text{GL}_n (\mathbb{C})$. I would like to use the decomposition $\mathfrak{g} = i \mathfrak{h} \oplus \mathfrak{h}$, where $\mathfrak{h}$ is the sub-$\mathbb{C}$-vector space of hermitian complex matrices and $i \mathfrak{h}$ is the sub-Lie algebra of skew-Hermitian matrices to achieve a decomposition $A = UP$ of any $A \in \text{GL}_n (\mathbb{C})$ into a unitary matrix $U$ and a positive definite matrix $P$, making use of the exponential map. However, there is an obvious obstruction: $\text{exp}(A + B)$ is not $\text{exp}(A) \text{exp}(B)$ necessarily. So the question is, how might we work around this problem?

Now, the subgroup $U \subset G$ of unitary matrices acts on the sub-$\mathbb{C}$-vector space $P \subset G$ of hermitian positive definite matrices by $(U, P) \mapsto U^* P U$, so this theorem will show that $G$ has a semi-direct product decomposition $G = U \times P$ with product $(U, P)(U', P') \mapsto (UU' , U'^* P U' P)$.

It is possibly useful to note that the map $i \mathfrak{h} \rightarrow \text{End}_{\mathbb{C}}(\mathfrak{h})$ sending $s$ to $\text{ad}_s$, where $\text{ad}_s (h) = sh - hs$ is a Lie-algebra representation.

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    $\begingroup$ just to make sure I understand the question: I always thought of the polar decomposition $A=UP$ as the matrix analogue of $z=|z| e^{i\,{\rm arg}\,z}$, so a decomposition into modulus and argument; then $P=(A^\ast A)^{1/2}$ and $U=AP^+$ (with $P^+$ the pseudo-inverse); this is not what you want? $\endgroup$ – Carlo Beenakker Oct 13 '18 at 12:05
  • $\begingroup$ Hi Carlo, that works. I like that since we don't have to use a norm. But I meant to ask something more specific: can the idea above be continued? Specifically, the idea is to use the decomposition of $\mathfrak{g}$ into $\mathfrak{h}$ and $i \mathfrak{h}$ and the exponential map. So, can we find a way to express $\text{exp} ( s+ h) = UP$ where $s$ is skew hermitian (i.e. $ih'$ for some hermitian operator $h$), $h$ is hermitian, $U$ is unitary, and $P$ is positive definite? As noted, not as simple as the case for $\mathbb{C}$, where $\text{exp}(z + w) = \text{exp}(z) \text{exp}(w)$. $\endgroup$ – Dean Young Oct 13 '18 at 13:49
  • $\begingroup$ Part of my interest is that I want the analogy with $\mathbb{C}$ to be more clearly expressed in the proof, but I don't know if it would work. $\endgroup$ – Dean Young Oct 13 '18 at 13:50
  • $\begingroup$ Actually, I think your method might lead to a way of saying just that - sorry. Maybe you can explain some of what you're thinking in an answer. $\endgroup$ – Dean Young Oct 13 '18 at 13:53
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    $\begingroup$ One source of the analogies mentioned at the outset is the classic book by Paul Halmos, Finite Dimensional Vector Spaces. $\endgroup$ – Jim Humphreys Oct 13 '18 at 14:49

I have always understood the polar decomposition as the matrix analogue of $z=|z|e^{i\,{\rm arg}z}$ so a decomposition into modulus and argument; the OP want the matrix analogue of a decomposition into real and imaginary parts, $z=e^{{\rm Re}\,z}e^{i\,{\rm Im}\,z}$, which I have never encountered and may not exist for the reason mentioned by the OP ($e^{X+iY}\neq e^Xe^{iY}$).

In any case, if we do follow the first approach, then $P=(A^∗A)^{1/2}$ and $U=AP^+$ (with $P^+$ the pseudo-inverse) gives the unique polar decomposition $A=UP$, with $P$ Hermitian positive semidefinite and $U^+=U^\ast$.

Generalizations of this polar decomposition exist (notably for non-square matrices, see The Canonical Generalized Polar Decomposition, 2009), but these generalizations also follow the modulus-argument decomposition (and not the real-imaginary decomposition).

  • $\begingroup$ Wait, it should be, "set $z = e^w$. Then $z = e^{\text{Re}(w)} e^{i \text{Im}(w)} = $". In that case, $e^{\text{Re}(w)} = |z|$ and $\text{Im}(w) = \text{arg}(z) \text{ mod } 2 \pi$ $\endgroup$ – Dean Young Oct 13 '18 at 14:21

The basic answer to the question here is "Yes, there is a strong analogy via the Iwasawa decomposition for a semisimple Lie group". If I were trying to study this kind of question, I'd probably start with a MathSciNet search, which of course does require access. For example, searching for "lie group" and "polar decomposition" (written this way in two Anywhere boxes) returns about 38 items with authors such as Kostant along with many Russians. Browsing through such a list gives some insight into the historical development.

Since this is pretty far from my actual specialization, I can't comment in more detail on what you might find out. But the main point is that Math Overflow might not be the place to start.


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