Invertible components and exponential laws

Written by GPT-6.1 Sol (OpenAI), Ultra, September–October 2026. Self-checked by the writing AI. Original text: CC0 1.0.

An invertible element can move continuously without losing its inverse. The connected components of the invertible group record the obstructions to moving it to the identity. Exponentials describe the identity component; in C*-algebras the same components can be studied using unitaries. For continuous functions on a circle, the obstruction becomes an integer winding number.

Prerequisites are Banach algebras, spectrum and holomorphic functional calculus and C*-algebras and continuous functional calculus. We use completeness, the Neumann series and the power-series definitions of exponential and logarithm. Integrals of norm-continuous Banach-space-valued functions over compact intervals can be defined as limits of Riemann sums. All paths and connected components below use the norm topology. Freely readable treatments are Blackadar’s Operator Algebras and Sundar’s Notes on C*-algebras.

1. The identity component is generated by exponentials

Let $A$ be a unital complex Banach algebra and let $G(A)$ be its group of invertible elements. Inversion is continuous: if $x\in G(A)$ and $\|x^{-1}(y-x)\|<1$, the Neumann series inverts $y=x(1+x^{-1}(y-x))$. Write $G_0(A)$ for the connected component of $1$.

Recall 1.1. The group $G_0(A)$ consists exactly of the finite products $$\exp(a_1)\exp(a_2)\cdots\exp(a_N),\qquad a_j\in A. \tag{1.1}$$ It is an open normal subgroup, and each of its elements can be joined to $1$ by a norm-continuous path of invertibles. Every connected component of $G(A)$ is a coset of $G_0(A)$.

This is Proposition 7.1(2), (5) of Banach algebras, spectrum and holomorphic functional calculus, where the complete proof is given. The coset assertion follows by translation of connected components. We will use this result both for the component group and for lifting invertibles from a quotient.

The component group $G(A)/G_0(A)$ can be noncommutative. We do not assume commutativity when using it.

Size of a local logarithm. If $r=\|1-x\|<1$, the norm-convergent logarithm series in the prerequisite gives $$\|\log x\|\leq\sum_{n=1}^\infty\frac{r^n}{n} =-\log(1-r).$$ The bound depends on $r$. There is no uniform bound on $\|\log x\|$ for all elements satisfying $\|1-x\|<1$: in $A=\mathbb C$, the principal logarithm of $x=\varepsilon$, $0<\varepsilon<1$, has absolute value $-\log\varepsilon$. Thus the logarithm existence statement in [Blackadar, II.1.5.3] is useful here, while the accompanying uniform $\pi/2$ bound in the accessed version cannot be used.

2. Norm-continuous one-parameter groups

Theorem 2.1. If $x:\mathbb R\to G(A)$ is a norm-continuous homomorphism, there is a unique $a\in A$ such that $$x(t)=\exp(ta)\quad(t\in\mathbb R), \qquad a=\lim_{t\to0}\frac{x(t)-1}{t}. \tag{2.1}$$

Proof. Choose $\delta>0$ so small that $$B=\int_0^\delta x(s)\,ds$$ is invertible: $B/\delta$ is arbitrarily close to $1$ as $\delta\downarrow0$. The group law gives $$(x(t)-1)B =\int_\delta^{\delta+t}x(s)\,ds-\int_0^t x(s)\,ds.$$ Dividing by $t$ and using norm continuity, for positive or negative $t$, shows that the limit in (2.1) exists and equals $$a=(x(\delta)-1)B^{-1}.$$ The group law now gives $x'(t)=x(t)a$. Since $x(t)$ commutes with every $x(s)$, it commutes with their difference quotients and their limit $a$. Differentiating $\exp(-ta)x(t)$ therefore gives zero. Its value at $t=0$ is $1$, so $x(t)=\exp(ta)$. Uniqueness follows by differentiation at zero. $\square$

The norm hypothesis explains why the generator is bounded. Strongly continuous groups on Hilbert space can have unbounded generators.

3. Two exponential approximation formulas

The commutator of $a,b\in A$ is $[a,b]=ab-ba$.

For these estimates we may work with a Banach-algebra norm for which $\|1\|=1$. This does not restrict the theorem. For nonzero $A$, if the original norm has a different identity norm, define the left-multiplication norm $$\|a\|_L=\sup_{\|x\|\leq1}\|ax\|.$$ Submultiplicativity gives $\|a\|_L\leq\|a\|$; using $x=1/\|1\|$ gives $\|a\|\leq\|1\|\|a\|_L$. Hence the norms are equivalent, and the new norm is complete. Also $\|ab\|_L\leq\|a\|_L\|b\|_L$ and $\|1\|_L=1$. They give the same convergent series, norm limits, paths and connected components. In the zero algebra all formulas are immediate. We use the normalized norm in the following proof and then transfer its limits back to the given norm.

Theorem 3.1. For arbitrary $a,b\in A$, $$\exp(a+b)=\lim_{n\to\infty}\bigl(\exp(a/n)\exp(b/n)\bigr)^n, \tag{3.1}$$ and $$\exp([a,b]) =\lim_{n\to\infty} \bigl(\exp(-a/n)\exp(-b/n)\exp(a/n)\exp(b/n)\bigr)^{n^2}. \tag{3.2}$$ Both limits are in norm.

Proof. We first record an estimate. For elements $u,v$ and an integer $m\geq1$, $$u^m-v^m=\sum_{j=0}^{m-1}u^{m-1-j}(u-v)v^j, \qquad \|u^m-v^m\|\leq m\|u-v\|\max(\|u\|,\|v\|)^{m-1}. \tag{3.3}$$ No commutativity is needed; the sum telescopes.

The exponential series gives, for fixed $a,b$, $$u_n=\exp(a/n)\exp(b/n)=1+(a+b)/n+O(n^{-2}),$$ while $v_n=\exp((a+b)/n)$ has the same first two terms. Their difference is $O(n^{-2})$, and their norms are at most $\exp(C/n)$ for a fixed $C$. Taking $m=n$ in (3.3) gives $\|u_n^n-v_n^n\|=O(n^{-1})$. Since $v_n^n=\exp(a+b)$, (3.1) follows.

For (3.2), set $$w(t)=\exp(-ta)\exp(-tb)\exp(ta)\exp(tb).$$ Multiplication of the four series through degree two yields $$w(t)=1+t^2(ab-ba)+O(|t|^3). \tag{3.4}$$ To justify the remainder, each exponential remainder after degree two is bounded by $|t|^3\|a\|^3e^{|t|\|a\|}/6$, or its $b$-analogue. Multiplying the four finite quadratic parts leaves finitely many terms of degree at least three, also $O(|t|^3)$ for $|t|\leq1$. Thus the estimate holds in an arbitrary Banach algebra. In particular $\|w(1/n)\|\leq1+C/n^2$, a sharper bound than estimating the four factors separately. Compare it with $z_n=\exp([a,b]/n^2)$. Their difference is $O(n^{-3})$, and both norms are at most $1+C'/n^2$. Taking $m=n^2$ in (3.3) gives a difference $O(n^{-1})$, while $z_n^{n^2}=\exp([a,b])$. $\square$

The quadratic bound on $w(1/n)$ is essential: a bound of the form $e^{C/n}$, raised to $n^2$, would grow with $n$ and would not prove convergence.

4. Unitaries carry the same components

Now let $A$ be a unital C*-algebra. Write $U(A)$ for its unitary group and $U_0(A)$ for the connected component of $1$ in that group.

Theorem 4.1. For $x\in G(A)$, the element $$u(x)=x|x|^{-1},\qquad |x|=(x^*x)^{1/2}, \tag{4.1}$$ is unitary, and depends continuously on $x$. Moreover $$G(A)=U(A)G_0(A),\qquad U(A)\cap G_0(A)=U_0(A), \tag{4.2}$$ and inclusion induces a group isomorphism $$U(A)/U_0(A)\cong G(A)/G_0(A). \tag{4.3}$$

Proof. In the zero algebra all assertions are immediate. Otherwise, since $x$ is invertible, $x^*x\geq\|x^{-1}\|^{-2}1$; continuous functional calculus makes $|x|$ invertible. Direct multiplication gives $u(x)^*u(x)=1$. The product $x|x|^{-1}$ is invertible, so its inverse is its adjoint and $u(x)u(x)^*=1$. The square-root map is norm-continuous on positive elements: on any fixed bounded spectral interval it is uniformly approximable by polynomials, and polynomial evaluation is continuous. Continuity of inversion proves continuity of (4.1).

Every positive invertible $h$ has a continuous-functional-calculus logarithm, so $h=\exp(\log h)\in G_0(A)$. The decomposition $x=u(x)|x|$ proves the first part of (4.2).

The unitary group is locally path connected. If a unitary $v$ is sufficiently close to $1$, its spectrum is in an arc admitting a continuous argument, and functional calculus gives $v=\exp(ik)$ for self-adjoint $k$; $t\mapsto\exp(itk)$ is a unitary path. Translation gives the same assertion near any unitary. Hence connected components of $U(A)$ are path components. If $u\in U(A)\cap G_0(A)$, Recall 1.1 gives an invertible path from $1$ to $u$; applying the continuous map $u(\cdot)$ to that path gives a unitary path with the same endpoints. This proves $U(A)\cap G_0(A)\subseteq U_0(A)$, and the reverse inclusion follows from any unitary path. The quotient map from $U(A)$ to $G(A)/G_0(A)$ is a homomorphism, surjective by the first part of (4.2), with kernel $U_0(A)$ by the second. This proves (4.3). $\square$

Although $u(\cdot)$ is useful for paths, it is generally not a group homomorphism. The quotient isomorphism uses the inclusion of unitaries, which is a homomorphism.

A continuous deformation to the unitary group

Adapted and expanded from S. Sundar [Sundar], Section 4.2, the remark following the opening proposition defining the stabilized component group. This entire subsection is CC0. AI changes by GPT-6.1 Sol (OpenAI), Ultra: joint continuity and the fixed-unitary property are proved explicitly; stabilization is not needed for this assertion.

Corollary 4.2. For every unital C*-algebra, the maps $$\begin{gathered} r_s(x)=x|x|^{-s}=x\exp(-s\log|x|),\\ 0\leq s\leq1,\quad x\in G(A), \end{gathered}$$ give a strong deformation retraction of $G(A)$ onto $U(A)$: the map $(s,x)\mapsto r_s(x)$ is continuous, $r_0(x)=x$, $r_1(x)=u(x)$, and $r_s(v)=v$ for every unitary $v$.

Proof. The zero algebra gives a constant deformation on a singleton. Otherwise all factors are invertible. Near a fixed invertible $x$, the spectra of $|x|$ lie in one compact interval $[\delta,M]\subset(0,\infty)$: continuity of $|x|$ gives the upper bound, and continuity of inversion gives a positive lower bound. Uniform polynomial approximation to the scalar logarithm on that interval, followed by continuous polynomial evaluation, proves continuity of $x\mapsto\log|x|$. The norm-convergent exponential series is uniformly convergent on bounded sets, so $(s,x)\mapsto\exp(-s\log|x|)$ is continuous. Multiplication proves joint continuity of $r_s$. The endpoints follow from functional calculus. For a unitary $v$, $|v|=1$, so every $r_s(v)$ equals $v$. $\square$

5. Winding on a circle

For $A=C(S^1)$, unitaries are continuous functions $u:S^1\to S^1$.

Lemma 5.1. Every continuous $g:\mathbb R\to S^1$ has a continuous lift $f:\mathbb R\to\mathbb R$ with $g(t)=e^{2\pi if(t)}$. Two lifts differ by a constant integer.

Proof. On a compact interval, uniform continuity gives a finite partition such that $g(t)/g(t_j)$ stays in a small arc around $1$ for $t\in[t_j,t_{j+1}]$. On this arc choose the continuous argument that is zero at $1$. Once $f(t_j)$ is chosen, set $$f(t)=f(t_j)+\frac1{2\pi}\operatorname{Arg}\bigl(g(t)/g(t_j)\bigr)$$ on that subinterval. Successive choices agree at their shared endpoints. Starting from a chosen lift of $g(0)$, apply this construction on $[k,k+1]$ successively for all nonnegative integers $k$, and on $[-k-1,-k]$ backwards for the negative side. The resulting lift is continuous on all of $\mathbb R$. The difference of two lifts is continuous and integer-valued, hence constant. $\square$

Apply the lemma to $g(t)=u(e^{2\pi it})$. For any lift $f_u$, define $$\deg u=f_u(1)-f_u(0)\in\mathbb Z. \tag{5.1}$$

Theorem 5.2. Degree is a surjective homomorphism $U(C(S^1))\to\mathbb Z$, with kernel $U_0(C(S^1))$. Consequently $$G(C(S^1))/G_0(C(S^1))\cong\mathbb Z. \tag{5.2}$$

Proof. The difference $f_u(t+1)-f_u(t)$ is continuous and integer-valued, so is constant. At $t=0$ it is (5.1), an integer since $g(1)=g(0)$. Changing a lift by a constant integer does not change degree. Adding lifts for $u,v$ gives a lift for $uv$; hence $\deg(uv)=\deg u+\deg v$. The functions $u(z)=z^k$ have degree $k$, proving surjectivity.

If $\deg u=0$, the lift $f_u$ is one-periodic and descends to a continuous real function $h$ on $S^1$. The path $u_s(z)=e^{2\pi ish(z)}$, for $0\leq s\leq1$, joins $1$ to $u$. Conversely, degree is locally constant in the uniform norm: if $v$ is sufficiently close to $u$, the unitary $vu^{-1}$ has values in an arc around $1$ and has a single-valued continuous real logarithm on the circle, so has degree zero. The homomorphism law gives $\deg v=\deg u$. It follows that degree is constant along any unitary path, and therefore vanishes on $U_0$. The kernel assertion and Theorem 4.1 prove (5.2). $\square$

6. Exercises with complete solutions

Exercise 6.1 — A generator from one short interval (intermediate). Let $x:\mathbb R\to G(A)$ be norm continuous and multiplicative. If $\|x(s)-1\|\leq\eta<1$ for $0\leq s\leq\delta$, prove that the generator satisfies $$\|a\|\leq\frac{\|x(\delta)-1\|}{\delta(1-\eta)}.$$

Solution. Put $e=1-B/\delta$ and $c=x(\delta)-1$. The average in Theorem 2.1 gives $\|e\|\leq\eta$. The Neumann series and the formula for the generator yield $$a=\frac1\delta c(1-e)^{-1} =\frac1\delta\sum_{n=0}^\infty ce^n.$$ The first summand is $c$; for $n\geq1$, submultiplicativity gives $\|ce^n\|\leq\|c\|\eta^n$. Summing proves the stated bound in the original norm, even if $\|1\|>1$.

Exercise 6.2 — Checking the commutator sign (basic). In $M_2(\mathbb C)$, put $a=e_{12},b=e_{21}$. Compute the first nonzero term of $\exp(-ta)\exp(-tb)\exp(ta)\exp(tb)$, and the limit in (3.2).

Solution. Here $a^2=b^2=0$, so all four exponentials equal $1\pm ta$ or $1\pm tb$. Multiplication gives $1+t^2(e_{11}-e_{22})+O(t^3)$. Thus $[a,b]=\operatorname{diag}(1,-1)$, and the limit is $\operatorname{diag}(e,e^{-1})$. Reversing $a,b$ reverses the sign and exchanges these two entries.

Exercise 6.3 — A circle with a harmless oscillation (intermediate). For $k\in\mathbb Z$ and real $c\neq0$, consider $u(e^{2\pi it})=\exp(2\pi i[kt+c\sin(2\pi t)])$. Determine its component in $U(C(S^1))$, and produce a path to $z\mapsto z^k$. Can it be joined to $1$ when $k\neq0$?

Solution. The displayed bracket is a lift; its endpoint difference is $k$. The path $u_s(e^{2\pi it})=\exp(2\pi i[kt+(1-s)c\sin(2\pi t)])$ is well-defined on the circle and joins $u$ to $z^k$. Its degree stays $k$. If $k\neq0$, Theorem 5.2 prohibits a path to $1$.

References

[Blackadar] Bruce Blackadar, Operator Algebras: Theory of C-Algebras and von Neumann Algebras*, author's revised and corrected online version of the 2005 book, accessed 3 October 2026.

[Sundar] S. Sundar, Notes on C*-algebras, arXiv:2505.17456v1, 23 May 2025, opening of Section 4.2 and its polar-path remark. Author-supplied TeX; CC0 licence.

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