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Robert Bryant
Robert Bryant
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The answer is 'no'. For example, let $$ X = \{\ B\otimes C\ | \det(B)\det(C) = 1\ \}\subset \mathrm{SL}(4,\mathbb{C}). $$ Then $X$ is isomorphic to the group $\mathrm{SO}(4,\mathbb{C})$ and hence $\pi_1(X)\simeq \mathbb{Z}_2$, while the set $$ G = \{\ (B, C)\ |\ \det(B)\det(C) = 1\ \}\subset \mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C}) $$ is a connected Lie group that satisfies $\pi_1(G)\simeq \mathbb{Z}$, with generator given by the subgroup $$ \{\ (B, B^{-1})\ |\ B\in T\ \} $$ where $T\subset\mathbb{GL}(2,\mathbb{C})$ is a circle subgroup that generates $\pi_1\bigl(\mathbb{GL}(2,\mathbb{C})\bigr)\simeq \mathbb{Z}$.

Now, the map $\tau:\mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C})\to \mathrm{GL}(4,\mathbb{C})$ given by $\tau(B,C) = B\otimes C$ is a group homomorphism, and the preimage under $\tau$ of the space $X$ is the two-component subgroup $$ G^+ = \{\ (B, C)\ |\ \det(B)\det(C) = \pm1\ \}\subset \mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C}). $$ (This is because of the identity $\det(B\otimes C) = \det(B)^2\det(C)^2$.) Note that $\pi_1(G^+) = \pi_1(G) = \mathbb{Z}_2$.

The map $\tau:G\to X$ given by $\tau(B,C) = B\otimes C$ is a group homomorphism. If there were a continuous map $\sigma:X\to G$$\sigma = (g,h):X\to G^+$ that satisfied $\tau\circ\sigma = \mathrm{id}_X$, this would induce a homomorphism $\sigma_*:\pi_1(X)\to\pi_1(G)$$\sigma_*:\pi_1(X)\to\pi_1(G^+)$ that satisfied $$ \tau_*\circ\sigma_* = \mathrm{id}:\pi_1(X)\to\pi_1(X)\simeq\mathbb{Z}_2. $$ However, then $\sigma_*$ would be a nontrival homomorphism $\mathbb{Z}_2\to\mathbb{Z}$, and this does not exist.

The answer is 'no'. For example, let $$ X = \{\ B\otimes C\ | \det(B)\det(C) = 1\ \}\subset \mathrm{SL}(4,\mathbb{C}). $$ Then $X$ is isomorphic to the group $\mathrm{SO}(4,\mathbb{C})$ and hence $\pi_1(X)\simeq \mathbb{Z}_2$, while the set $$ G = \{\ (B, C)\ |\ \det(B)\det(C) = 1\ \}\subset \mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C}) $$ is a connected Lie group that satisfies $\pi_1(G)\simeq \mathbb{Z}$, with generator given by the subgroup $$ \{\ (B, B^{-1})\ |\ B\in T\ \} $$ where $T\subset\mathbb{GL}(2,\mathbb{C})$ is a circle subgroup that generates $\pi_1\bigl(\mathbb{GL}(2,\mathbb{C})\bigr)\simeq \mathbb{Z}$.

The map $\tau:G\to X$ given by $\tau(B,C) = B\otimes C$ is a group homomorphism. If there were a continuous map $\sigma:X\to G$ that satisfied $\tau\circ\sigma = \mathrm{id}_X$, this would induce a homomorphism $\sigma_*:\pi_1(X)\to\pi_1(G)$ that satisfied $$ \tau_*\circ\sigma_* = \mathrm{id}:\pi_1(X)\to\pi_1(X)\simeq\mathbb{Z}_2. $$ However, then $\sigma_*$ would be a nontrival homomorphism $\mathbb{Z}_2\to\mathbb{Z}$, and this does not exist.

The answer is 'no'. For example, let $$ X = \{\ B\otimes C\ | \det(B)\det(C) = 1\ \}\subset \mathrm{SL}(4,\mathbb{C}). $$ Then $X$ is isomorphic to the group $\mathrm{SO}(4,\mathbb{C})$ and hence $\pi_1(X)\simeq \mathbb{Z}_2$, while the set $$ G = \{\ (B, C)\ |\ \det(B)\det(C) = 1\ \}\subset \mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C}) $$ is a connected Lie group that satisfies $\pi_1(G)\simeq \mathbb{Z}$, with generator given by the subgroup $$ \{\ (B, B^{-1})\ |\ B\in T\ \} $$ where $T\subset\mathbb{GL}(2,\mathbb{C})$ is a circle subgroup that generates $\pi_1\bigl(\mathbb{GL}(2,\mathbb{C})\bigr)\simeq \mathbb{Z}$.

Now, the map $\tau:\mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C})\to \mathrm{GL}(4,\mathbb{C})$ given by $\tau(B,C) = B\otimes C$ is a group homomorphism, and the preimage under $\tau$ of the space $X$ is the two-component subgroup $$ G^+ = \{\ (B, C)\ |\ \det(B)\det(C) = \pm1\ \}\subset \mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C}). $$ (This is because of the identity $\det(B\otimes C) = \det(B)^2\det(C)^2$.) Note that $\pi_1(G^+) = \pi_1(G) = \mathbb{Z}_2$.

The map $\tau:G\to X$ given by $\tau(B,C) = B\otimes C$ is a group homomorphism. If there were a continuous map $\sigma = (g,h):X\to G^+$ that satisfied $\tau\circ\sigma = \mathrm{id}_X$, this would induce a homomorphism $\sigma_*:\pi_1(X)\to\pi_1(G^+)$ that satisfied $$ \tau_*\circ\sigma_* = \mathrm{id}:\pi_1(X)\to\pi_1(X)\simeq\mathbb{Z}_2. $$ However, then $\sigma_*$ would be a nontrival homomorphism $\mathbb{Z}_2\to\mathbb{Z}$, and this does not exist.

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Robert Bryant
Robert Bryant
  • 108.4k
  • 8
  • 341
  • 453

The answer is 'no'. For example, let $$ X = \{\ B\otimes C\ | \det(B)\det(C) = 1\ \}\subset \mathrm{SL}(4,\mathbb{C}). $$ Then $X$ is isomorphic to the group $\mathrm{SO}(4,\mathbb{C})$ and hence $\pi_1(X)\simeq \mathbb{Z}_2$, while the set $$ G = \{\ (B, C)\ |\ \det(B)\det(C) = 1\ \}\subset \mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C}) $$ is a connected Lie group that satisfies $\pi_1(G)\simeq \mathbb{Z}$, with generator given by the subgroup $$ \{\ (B, B^{-1})\ |\ B\in T\ \} $$ where $T\subset\mathbb{GL}(2,\mathbb{C})$ is a circle subgroup that generates $\pi_1\bigl(\mathbb{GL}(2,\mathbb{C})\bigr)\simeq \mathbb{Z}$.

The map $\tau:G\to X$ given by $\tau(B,C) = B\otimes C$ is a group homomorphism. If there were a continuous map $\sigma:X\to G$ that satisfied $\tau\circ\sigma = \mathrm{id}_X$, this would induce a homomorphism $\sigma_*:\pi_1(X)\to\pi_1(G)$ that satisfied $$ \tau_*\circ\sigma_* = \mathrm{id}:\pi_1(X)\to\pi_1(X)\simeq\mathbb{Z}_2. $$ However, then $\sigma_*$ would be a nontrival homomorphism $\mathbb{Z}_2\to\mathbb{Z}$, and this does not exist.