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Exercise 1376Topology › Topology of Mn(ℝ) › Exponential, logarithm and group topologyL3Added 6 Nov 2026

The logarithm near the identity and local inverse of exp

#matrix logarithm#local homeomorphism#power series

Statement

Here ∥⋅∥\|\cdot\| is the operator norm. For ∥X∥<1\|X\|<1 define log⁡(I+X)=∑k≥1(−1)k+1kXk\log(I+X)=\sum_{k\ge1}\frac{(-1)^{k+1}}kX^k. Show that:

  1. the series converges absolutely;
  2. elog⁡(I+X)=I+Xe^{\log(I+X)}=I+X for ∥X∥<1\|X\|<1, and log⁡(eA)=A\log(e^A)=A for ∥A∥<ln⁡2\|A\|<\ln2;
  3. exp⁡\exp restricts to a homeomorphism from the ball B(0,ln⁡2)B(0,\ln2) onto an open set containing B(I,12)B(I,\tfrac12), with inverse log⁡\log.
Hint

Use the formal power series identities exp⁡(log⁡(1+x))=1+x\exp(\log(1+x))=1+x and log⁡(1+(ex−1))=x\log(1+(e^x-1))=x, and justify substitution by absolute convergence.

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Solution

1. ∑k∥X∥kk=−ln⁡(1−∥X∥)<∞\sum_k\frac{\|X\|^k}k=-\ln(1-\|X\|)<\infty for ∥X∥<1\|X\|<1.

2. The identity elog⁡(I+X)=I+Xe^{\log(I+X)}=I+X. Let xx be a formal variable. In Q[[x]]\mathbb{Q}[[x]] one has exp⁡(log⁡(1+x))=1+x\exp(\log(1+x))=1+x (a known identity of formal power series). Substituting XX for xx is legitimate when the double series of norms converges: the norm-majorant is
∑m≥01m!(∑k≥1∥X∥kk)m=exp⁡(−ln⁡(1−∥X∥))=11−∥X∥<∞.\sum_{m\ge0}\frac1{m!}\Bigl(\sum_{k\ge1}\frac{\|X\|^k}k\Bigr)^m=\exp\bigl(-\ln(1-\|X\|)\bigr)=\frac1{1-\|X\|}<\infty.
Absolute convergence allows rearranging the terms into powers of XX, and the coefficients are those of the formal identity. Hence elog⁡(I+X)=I+Xe^{\log(I+X)}=I+X.

Similarly, for ∥A∥<ln⁡2\|A\|<\ln2 put X=eA−IX=e^A-I. Then ∥X∥≤e∥A∥−1<1\|X\|\le e^{\|A\|}-1<1 and the formal identity log⁡(exp⁡x)=x\log(\exp x)=x (that is, log⁡(1+(ex−1))=x\log(1+(e^x-1))=x) gives log⁡(eA)=A\log(e^A)=A once the double series converges absolutely: its majorant is
∑k(e∥A∥−1)kk=−ln⁡(2−e∥A∥)<∞exactly when ∥A∥<ln⁡2.\sum_k\frac{(e^{\|A\|}-1)^k}k=-\ln\bigl(2-e^{\|A\|}\bigr)<\infty\quad\text{exactly when }\|A\|<\ln2.

3. Local homeomorphism. Let U={I+X:∥X∥<1, ∥log⁡(I+X)∥<ln⁡2}U=\{I+X:\|X\|<1,\ \|\log(I+X)\|<\ln2\}. It is open, as the preimage of B(0,ln⁡2)B(0,\ln2) under the continuous map log⁡\log on the open ball B(I,1)B(I,1) (intersected with that ball).
- If ∥A∥<ln⁡2\|A\|<\ln2, then ∥eA−I∥<1\|e^A-I\|<1 and log⁡eA=A\log e^A=A has norm <ln⁡2<\ln2, so eA∈Ue^A\in U.
- If I+X∈UI+X\in U then A=log⁡(I+X)∈B(0,ln⁡2)A=\log(I+X)\in B(0,\ln2) and eA=I+Xe^A=I+X.

So exp⁡:B(0,ln⁡2)→U\exp:B(0,\ln2)\to U is a bijection with continuous inverse log⁡\log, hence a homeomorphism. Finally U⊃B(I,12)U\supset B(I,\frac12): if ∥X∥<12\|X\|<\frac12 then ∥log⁡(I+X)∥≤−ln⁡(1−∥X∥)<ln⁡2\|\log(I+X)\|\le-\ln(1-\|X\|)<\ln2.

Remark. This shows that exp⁡\exp is a local homeomorphism at 00: every matrix close to II has a (unique small) logarithm. It is the first step in proving that closed subgroups of GLnGL_n are Lie groups.