[proofplan] We show $G(\mathcal{L}(H))$ is path-connected with $I = \mathrm{id}_H$ as a star point: every invertible $T$ is connected to $I$ by a continuous path inside $G(\mathcal{L}(H))$. The key ingredients are the [Polar Decomposition of Invertible Operators](/theorems/2691), which writes $T = RU$ with $R$ positive invertible and $U$ unitary, and two exponential representations: a real exponential $R = e^S$ with $S$ self-adjoint (built from $\log$ via the Borel Functional Calculus on the strictly positive spectrum of $R$), and the **Representation of Unitary Operators** $U = e^{iQ}$ with $Q$ self-adjoint. The path $\gamma(t) = e^{tS} e^{itQ}$ then connects $I$ to $T$ inside $G(\mathcal{L}(H))$, and continuity of the operator exponential closes the argument. [/proofplan]

[step:Reduce connectedness to the existence of a continuous path from $I$ to each $T \in G(\mathcal{L}(H))$] Path-connectedness implies connectedness in any topological space. We show that for every $T \in G(\mathcal{L}(H))$ there is a continuous map \begin{align*} \gamma_T: [0, 1] \to G(\mathcal{L}(H)) \end{align*} with $\gamma_T(0) = I$ and $\gamma_T(1) = T$. Once this is established, $G(\mathcal{L}(H))$ is path-connected (any two points $T_0, T_1$ are joined via $I$ by concatenating $\gamma_{T_0}$ run in reverse with $\gamma_{T_1}$), and hence connected. We fix $T \in G(\mathcal{L}(H))$ and construct $\gamma_T$ in the next steps. [/step]

[step:Apply polar decomposition to write $T = RU$ with $R$ positive invertible and $U$ unitary] By the [Polar Decomposition of Invertible Operators](/theorems/2691) applied to $T$ — whose hypothesis (invertibility of $T$) is given — there exist unique $R, U \in \mathcal{L}(H)$ such that: (a) $R$ is positive, (b) $U$ is unitary, (c) $T = RU$. Moreover, the existence proof there gives the explicit formula $R = (TT^*)^{1/2}$ and $U = R^{-1} T$. *$R$ is invertible.* Since $T$ is invertible, $T^*$ is invertible (its inverse is $(T^{-1})^*$), and the product $TT^*$ is invertible (product of invertibles). The square root of an invertible positive operator is invertible: indeed, $R^2 = TT^*$ is invertible, so $R$ has a left inverse $R^{-1} = R \cdot (R^2)^{-1} = R \cdot (TT^*)^{-1}$ — and a right inverse by symmetry, so $R$ is invertible. Equivalently: $R$ is positive and invertible iff $\sigma(R) \subseteq (0, \infty)$. Since $R^2 = TT^*$ is invertible, $0 \notin \sigma(R^2)$. By the spectral mapping theorem for the polynomial $p(\lambda) = \lambda^2$ (or directly from the [Spectrum is Compact](/theorems/2649) and the BFC), $\sigma(R^2) = \{\lambda^2 : \lambda \in \sigma(R)\}$, hence $0 \notin \sigma(R)$, hence $\sigma(R) \subseteq (0, \infty)$ since $R$ is positive. Set $K_R := \sigma(R) \subseteq (0, \infty)$, which is non-empty and compact. [/step]

[step:Construct a self-adjoint $S \in \mathcal{L}(H)$ with $e^S = R$ via $\log$] We define $S := \log(R)$ using the [Borel Functional Calculus](/theorems/2696) for the normal (in fact self-adjoint) operator $R$, then verify $e^S = R$ by the same uniform-approximation argument used for the unitary representation. *The function $\log$ is bounded and continuous on $K_R$.* The natural logarithm $\log: (0, \infty) \to \mathbb{R}$ is continuous. Since $K_R \subseteq (0, \infty)$ is compact, $K_R \subseteq [a, b]$ for some $0 < a \le b < \infty$ (with $a = \min K_R > 0$, $b = \max K_R < \infty$, both attained on the compact set $K_R$). On $[a, b]$, $\log$ is continuous, hence bounded: $\|\log\|_{L^\infty(K_R)} \le \max(|\log a|, |\log b|) < \infty$. Hence $\log \in C(K_R) \subseteq L^\infty(K_R, \mathcal{B}(K_R))$. *Define $S$.* By the [Borel Functional Calculus](/theorems/2696) for the normal operator $R$ (every self-adjoint operator is normal: $R R^* = R^2 = R^* R$), let \begin{align*} \Psi_R: L^\infty(K_R, \mathcal{B}(K_R)) \to \mathcal{L}(H) \end{align*} be the associated unital $*$-homomorphism. Define \begin{align*} S := \Psi_R(\log) = \log(R) \in \mathcal{L}(H). \end{align*} *$S$ is self-adjoint.* Since $\log$ is real-valued on $(0, \infty)$, $\overline{\log} = \log$ pointwise on $K_R$. As $\Psi_R$ is a $*$-homomorphism, $S^* = \Psi_R(\overline{\log}) = \Psi_R(\log) = S$. *$e^S = R$.* By the same argument as in the proof of the **Representation of Unitary Operators**, the operator-norm exponential $e^S = \sum_{n=0}^\infty S^n/n!$ equals $\Psi_R(\exp \circ \log)$, where the polynomial truncations of the Taylor series of $\exp$ converge to $\exp$ uniformly on the bounded set $\log(K_R) \subseteq \mathbb{R}$ (recall $\log(K_R) \subseteq [\log a, \log b]$ from above, bounded), and $\Psi_R$ is bounded. Since $\exp(\log(\lambda)) = \lambda = \mathrm{id}(\lambda)$ for every $\lambda \in K_R \subseteq (0, \infty)$, we have $\exp \circ \log = \mathrm{id}$ in $L^\infty(K_R)$. Hence \begin{align*} e^S = \Psi_R(\exp \circ \log) = \Psi_R(\mathrm{id}) = \mathrm{id}(R) = R, \end{align*} using property (i) of the BFC. [claim:For any self-adjoint $A \in \mathcal{L}(H)$ with bounded spectrum lying in a Borel set on which a continuous function $h$ is defined, $h(A)$ defined via the BFC satisfies $e^{h(A)} = (\exp \circ h)(A)$ in operator norm] [proof] This is the same argument used in Step 4 of the **Representation of Unitary Operators** proof, applied here to $A = R$ and $h = \log$. The polynomial truncations $h_N := \sum_{n=0}^N h^n/n!$ converge to $\exp \circ h$ uniformly on the bounded set $h(K_A)$ (by the standard Taylor estimate $|\exp(z) - \sum_{n=0}^N z^n/n!| \le \sum_{n>N} M^n/n!$ for $|z| \le M$, taken with $M = \|h\|_{L^\infty(K_A)}$). Boundedness of the BFC (property (ii)) transports this uniform convergence to operator-norm convergence: \begin{align*} \Psi_A(h_N) \to \Psi_A(\exp \circ h) \quad \text{in } \mathcal{L}(H). \end{align*} The homomorphism property of $\Psi_A$ identifies $\Psi_A(h_N) = \sum_{n=0}^N h(A)^n/n!$, which is the partial sum of the operator exponential of $h(A)$. Hence the limit is $e^{h(A)}$, and $e^{h(A)} = \Psi_A(\exp \circ h) = (\exp \circ h)(A)$. [/proof] [/claim] Applying the claim to $A = R$, $h = \log$ gives $e^S = (\exp \circ \log)(R) = \mathrm{id}(R) = R$ as above. [/step]

[step:Apply the unitary representation to write $U = e^{iQ}$ with $Q$ self-adjoint] By the **Representation of Unitary Operators** (the immediately preceding theorem in this batch), applied to the unitary $U$ from Step 2, there exists a self-adjoint $Q \in \mathcal{L}(H)$ such that \begin{align*} U = e^{iQ}. \end{align*} The hypothesis "$U$ is unitary" is given by Step 2. Combining with Step 3, \begin{align*} T = RU = e^S \cdot e^{iQ}. \end{align*} [/step]

[step:Define the path $\gamma(t) := e^{tS} e^{itQ}$ and verify $\gamma(0) = I$, $\gamma(1) = T$] Define \begin{align*} \gamma: [0, 1] &\to \mathcal{L}(H) \\ t &\mapsto e^{tS} \cdot e^{itQ}. \end{align*} *$\gamma(0) = I$.* Since $0 \cdot S = 0$ and the operator exponential of $0$ is $I$ (every term except $n = 0$ vanishes), $e^{0 \cdot S} = I$ and $e^{0 \cdot iQ} = I$. Hence $\gamma(0) = I \cdot I = I$. *$\gamma(1) = T$.* By Step 4, $\gamma(1) = e^S \cdot e^{iQ} = R \cdot U = T$. [/step]

[step:Verify each $\gamma(t)$ is invertible] For each $t \in [0, 1]$, $tS$ and $itQ$ are bounded operators (scalar multiples of bounded operators), so the operator exponentials $e^{tS}$ and $e^{itQ}$ are well-defined and bounded. By the standard identity for the operator exponential of a single operator ($A$ commutes with itself, hence with $-A$), $e^A \cdot e^{-A} = e^{A - A} = e^0 = I$ for any $A \in \mathcal{L}(H)$. Applied to $A = tS$ and $A = itQ$: \begin{align*} e^{tS} \cdot e^{-tS} = I, \qquad e^{itQ} \cdot e^{-itQ} = I. \end{align*} Hence $e^{tS}$ and $e^{itQ}$ are invertible with inverses $e^{-tS}$ and $e^{-itQ}$. The product of invertible operators is invertible, so $\gamma(t) = e^{tS} \cdot e^{itQ}$ is invertible with \begin{align*} \gamma(t)^{-1} = (e^{itQ})^{-1} \cdot (e^{tS})^{-1} = e^{-itQ} \cdot e^{-tS}. \end{align*} Hence $\gamma(t) \in G(\mathcal{L}(H))$ for every $t \in [0, 1]$. [/step]

[step:Verify continuity of $\gamma: [0, 1]\to \mathcal{L}(H)$] We show $\gamma$ is continuous in the operator norm. It suffices to show that the two factors $t \mapsto e^{tS}$ and $t \mapsto e^{itQ}$ are continuous, since the product map $\mathcal{L}(H) \times \mathcal{L}(H) \to \mathcal{L}(H)$, $(A, B) \mapsto AB$, is jointly continuous on bounded sets: \begin{align*} \|A_1 B_1 - A_2 B_2\|_{\mathcal{L}(H)} \le \|A_1 - A_2\|_{\mathcal{L}(H)} \|B_1\|_{\mathcal{L}(H)} + \|A_2\|_{\mathcal{L}(H)} \|B_1 - B_2\|_{\mathcal{L}(H)}, \end{align*} and continuous functions on a compact interval take values in bounded sets. [claim:The map $\mathbb{R} \to \mathcal{L}(H)$, $t \mapsto e^{tA}$, is continuous for any $A \in \mathcal{L}(H)$] [proof] Fix $A \in \mathcal{L}(H)$. For $t, t' \in \mathbb{R}$, write \begin{align*} e^{tA} - e^{t'A} = e^{t'A} \cdot (e^{(t - t')A} - I) = e^{t'A} \sum_{n=1}^\infty \frac{((t-t')A)^n}{n!}. \end{align*} Hence \begin{align*} \|e^{tA} - e^{t'A}\|_{\mathcal{L}(H)} \le \|e^{t'A}\|_{\mathcal{L}(H)} \sum_{n=1}^\infty \frac{|t - t'|^n \|A\|^n}{n!} = \|e^{t'A}\|_{\mathcal{L}(H)} \cdot (e^{|t - t'| \|A\|} - 1). \end{align*} The factor $\|e^{t'A}\|_{\mathcal{L}(H)} \le e^{|t'| \|A\|}$ is bounded on bounded intervals, and $e^{|t - t'| \|A\|} - 1 \to 0$ as $t \to t'$. Hence $t \mapsto e^{tA}$ is continuous. [/proof] [/claim] Apply the claim with $A = S$ to obtain continuity of $t \mapsto e^{tS}$ on $[0, 1]$, and with $A = iQ$ to obtain continuity of $t \mapsto e^{itQ}$ on $[0, 1]$ (note $\|iQ\| = \|Q\|$, and the path is parametrised as $t \mapsto e^{t \cdot iQ}$, which fits the claim's form with $A = iQ$). By the joint continuity of multiplication on bounded sets, $\gamma(t) = e^{tS} e^{itQ}$ is continuous from $[0, 1]$ into $\mathcal{L}(H)$. Since each $\gamma(t)$ lies in $G(\mathcal{L}(H))$ (Step 6), $\gamma$ is continuous as a map $[0, 1] \to G(\mathcal{L}(H))$ when $G(\mathcal{L}(H))$ carries the subspace topology from $\mathcal{L}(H)$.[/step]

[guided]Why does continuity in $\mathcal{L}(H)$ suffice for continuity into $G(\mathcal{L}(H))$? Because $G(\mathcal{L}(H))$ carries the subspace topology: a map $[0, 1] \to G(\mathcal{L}(H))$ is continuous if and only if it is continuous as a map $[0, 1] \to \mathcal{L}(H)$ that happens to take values in $G(\mathcal{L}(H))$. The fact that $G(\mathcal{L}(H))$ is open in $\mathcal{L}(H)$ (Banach algebra fact: invertible elements form an open set) is not even needed here — the subspace topology argument is purely formal. The continuity claim itself is the content. We prove $t \mapsto e^{tA}$ is continuous for any bounded $A$, by writing the difference $e^{tA} - e^{t'A}$ as a multiplicative correction: \begin{align*} e^{tA} - e^{t'A} = e^{t'A}(e^{(t-t')A} - I). \end{align*} This factorisation works because $e^{tA} = e^{t'A + (t-t')A}$ and $t'A$ commutes with $(t-t')A$ (they are both scalar multiples of $A$, hence commute), so the exponential of a sum equals the product of exponentials in the commuting case. Then $e^{(t-t')A} - I = \sum_{n \ge 1} ((t-t')A)^n / n!$ has norm bounded by $e^{|t-t'| \|A\|} - 1 \to 0$ as $t \to t'$. Continuity of $\gamma(t) = e^{tS} e^{itQ}$ now follows from continuity of multiplication: a product of two continuous functions into a Banach algebra is continuous. We have used the standard estimate \begin{align*} \|A_1 B_1 - A_2 B_2\| \le \|A_1 - A_2\| \cdot \|B_1\| + \|A_2\| \cdot \|B_1 - B_2\|, \end{align*} which expresses the joint continuity of multiplication on bounded sets in any Banach algebra.[/guided]

[step:Wrap up — every invertible operator is path-connected to $I$] For arbitrary $T \in G(\mathcal{L}(H))$, the path $\gamma_T := \gamma$ constructed above (depending on the polar data $S$, $Q$ of $T$) is a continuous map $[0, 1] \to G(\mathcal{L}(H))$ with $\gamma_T(0) = I$ and $\gamma_T(1) = T$ (Step 5, 6, 7). Hence $G(\mathcal{L}(H))$ is star-shaped with star centre $I$: every point is connected to $I$ by a continuous path lying entirely inside $G(\mathcal{L}(H))$. Star-shaped subsets of any topological space are path-connected (concatenate the path from $T_0$ to $I$ — that is, $\gamma_{T_0}$ traversed in reverse, $t \mapsto \gamma_{T_0}(1 - t)$ — with the path from $I$ to $T_1$, namely $\gamma_{T_1}$). Path-connected spaces are connected. Therefore $G(\mathcal{L}(H))$ is connected. [/step]