Is my logic correct?

$f:U(n)\rightarrow U(1)$ defined by $f(A)=\det A$ is a group homomorphism so that the induced homomorphism $f^{*}: \pi_1(U(n))\rightarrow \pi_1(U(1))$ will be an isomorphism, right (I am not sure)? as $\pi_1(U(1))=\mathbb{Z}$ as $U(1)=S^1$ so $\pi_1(U(n))=\mathbb{Z}$.

  • 2
    $\begingroup$ Why is it an isomorphism? $\endgroup$ – user38268 Oct 27 '12 at 12:15
  • $\begingroup$ Well, I just got an counter example that it is not true, $\mathbb{R}\rightarrow S^1$, $x\mapsto e^{ix}$ $\endgroup$ – Marso Oct 27 '12 at 12:17
  • $\begingroup$ Well, it actually is an isomorphism in this case, but you're right that it isn't in general. $\endgroup$ – Jason DeVito Oct 27 '12 at 12:39
  • 3
    $\begingroup$ Have you tried using the long exact sequence of homotopy groups for $U(n-1)\hookrightarrow U(n)\to U(n)/U(n-1)$? $\endgroup$ – Neal Oct 27 '12 at 13:41

As you saw by your counter example, a Lie group homomorphism does not induce an isomorphism of fundamental groups. But one way to use homomorphisms to determine fundamental groups is through the fact that if $G$ and $H$ are connected with $G$ simply connected and $G \to H$ is a surjective homomorphism with a discrete kernel $K$ contained in the center of $G$, then this map is a covering and the fundamental group of $H$ is isomorphic to $K$. So if you know that $SU(n)$ is simply connected then you can consider the homomorphism $$ SU(n) \times \mathbb R \to U(n), ~~ (A, t) \mapsto e^{it} A. $$ This is surjective with kernel isomorphic to $\mathbb Z$ so that $\pi_1(U(n)) \simeq \mathbb Z$.

Though I guess the easiest way to see that $SU(n)$ is simply connected is to use Neal's suggestion of applying the LES in homotopy associated to the fibration $$SU(n-1) \to SU(n) \to SU(n)/SU(n-1) \simeq S^{2n-1}.$$

But then you might as well compute $\pi_1 U(n)$ directly from the similar fibration $$U(n-1) \to U(n) \to U(n)/U(n-1) \simeq S^{2n-1}.$$


The map $\mathrm{det}:U(n)\to U(1)\simeq S^1$ is a fiber bundle with fibers diffeomorphic to $\mathrm{SU}(n)$. Since $\mathrm{SU}(n)$ is connected and simply connected, as outlined in Eric Korman's answer, the long exact sequence associated to this fiber bundle provides an isomorphism $\pi_1(\mathrm{U}(n))\simeq \pi_1(S^1)\simeq \mathbb{Z}$.

  • $\begingroup$ This shows that in fact $\mathrm{det}_*$ is an isomorphism in this case. $\endgroup$ – F M Jan 12 '18 at 13:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.