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I'm interested in computing the fundamental group of the twisted loop space $$\Omega_f(M)=\{ \gamma \in C^{\infty}(\Bbb R,M) \mid \gamma(s+1)=f\gamma(s)\}$$ where $f \in \text{Aut}(M,x_0)$, for example a diffeomorphism with a fixed point $x_0$.

The twisted loop space is part of a fibration $$\Omega_{x_0}M \to \Omega_f(M) \xrightarrow{ev_0} M$$ where the map $ev_0$ is the evaluation map at $s=0$ and $\Omega_{x_0}M$ is the loop space (based at $x_0$).

The relevant piece of the l.e.s. homotopy gives $$\pi_2(M,x_0)\xrightarrow{\partial} \pi_1(\Omega_{x_0}M,x_0)\to \pi_1(\Omega_f(M),x_0) \to \pi_1(M,x_0)\to \pi_0(\Omega_{x_0}M)\to \pi_0(\Omega_f(M))$$

The last map should be an injection, hence we have a surjection $\pi_1(\Omega_f(M),x_0) \twoheadrightarrow \pi_1(M,x_0)$. I don't quite understand the effect of the boundary map from the second homotopy group of $M$, so I cannot really go on. As far as I know, if $M$ is simply connected then $\pi_1(\Omega_f(M),x_0) \cong \pi_2(M,x_0)$. If $\delta=0$ that would be the case but I cannot verify it.

Is this computation done somewhere so I can check how's done?

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  • $\begingroup$ Are you sure the last map is an injection? I would think $\pi_1(M, x_0 ) \to \pi_0 (\Omega_{x_0} M)$ is something like $\sigma \to f(\sigma) \sigma^{-1}$, which is not the zero map. $\endgroup$
    – Will Sawin
    Feb 25 at 22:32
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    $\begingroup$ Think about what happens when you take a loop in $\Sigma_f(M)$ starting at $x_0$ and ending at $x_0$ and flow it along a loop $\sigma$ in $M$. We can flow by simply moving the starting point along the map of $\sigma$ and the ending point along the path of $\gamma (\sigma)$, and extending the path to follow the new starting and ending points. The final result is our original path, composed with $\gamma(\sigma)$ on the left and $\sigma^{-1}$ on the right. This shows how $\pi_1$ acts on $\pi_0$, and gives several consequences: $\endgroup$
    – Will Sawin
    Feb 25 at 22:40
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    $\begingroup$ $\pi_0(\Omega_f(M))$ is the set of orbits in $\pi_1(M)$ of the $\gamma$-conjugacy action, i.e. $g_1$ and $g_2$ are in the same orbit if there exists $h$ with $ \gamma(h) g_1 h^{-1} =g_2$. The image of the map $\pi_1( \Omega_f(M), x_0) \to \pi_1(M, x_0)$ consists of only the $\gamma$-invariant elements of $\pi_1 (M , x_0)$ - it's not so hard to see how a homotopy between a loop representing an element of $\pi_1(M)$ and its image under $\Gamma$ gives a loop in $\Omega_f(M)$. $\endgroup$
    – Will Sawin
    Feb 25 at 22:45
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    $\begingroup$ The action is of $\pi_1(M, x_0)$ on $\pi_0 (\Omega_f(M))$ - probably a typo. The rule for this action is the same as any fibration - you take a map from $[0,1]$ to the base $M$ sending $0$ and $1$ to $x_0$, you lift the $0$ to a point in any fixed connected component of the fiber, then you lift the map from $[0,1]$, and you see what component $1$ ends up in. $\endgroup$
    – Will Sawin
    Feb 26 at 1:01
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    $\begingroup$ In this case we want a map from $[0,1]$ to a space of maps from $[0,1]$ to $M$, so we want a map from the square to $M$, and we know what the map should do on three of the sides, so we can just fill in by retracting the square to the union of those three sides. Sorry I meant $f(\sigma)$, the action of the automorphism $f$ on the loop $\sigma$. $\endgroup$
    – Will Sawin
    Feb 26 at 1:02
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Edit: The following is incorrect, see below. It might be more useful to think of $\Omega_f(M)$ as sitting in a homotopy pullback $\require{AMScd}$ \begin{CD} \Omega_f(M) @>>> M\\ @V V V @VVfV\\ M @>>\operatorname{Id}> M \end{CD}

Then you can make use of the "Mayer-Vietoris sequence" of homotopy groups $$\cdots \to \pi_2(M, x_0) \stackrel{\partial}{\to} \pi_1(\Omega_f(M),x_0) \to \pi_1(M, x_0) \times \pi_1(M, x_0) \to \pi_1(M,x_0) \to \cdots$$ described in the answers here. The last map above is $(a,b)\mapsto a\cdot f_*(b)^{-1}$, which is not a homomorphism in general but is a surjection of pointed sets. This immmediately shows for example that when $\pi_2(M,x_0)=0$ then $\pi_1(\Omega_f(M),x_0)$ is isomorphic to the graph of the homomorphism $f_*:\pi_1(M,x_0)\to \pi_1(M,x_0)$.

To go further you'd have to analyse the connecting map $\partial: \pi_2(M,x_0)\to \pi_1(\Omega_f(M),x_0)$. It may be (though I'm not sure) that the map preceding it is given by $(x,y)\mapsto x-f_*(y)$, in which case $\partial=0$. A good place to read up on this sequence seems to be May and Ponto's More concise algebraic topology, Section 2.2

Edit: as Tyrone points out, I wrote down the wrong pullback square, and consequently my conclusions were incorrect. What I should have written down was the topological pullback

$\require{AMScd}$ \begin{CD} \Omega_f(M) @>>> C^\infty(\mathbb{R},M)\\ @V V V @VV\operatorname{ev}_{0,1}V\\ M @>>(\operatorname{Id},f)> M\times M \end{CD} in which the map $\operatorname{ev}_{0,1}$ is homotopically equivalent to the diagonal map $\triangle: M\to M\times M$. Then the MV sequence looks like

$$\cdots \to \pi_2(M \times M) \stackrel{\partial}{\to} \pi_1(\Omega_f(M)) \to \pi_1(M \times M) \to \pi_1(M\times M) \to \cdots$$

where this time the last map is given by $(a,b)\mapsto(a b^{-1}, f_*(a) b^{-1})$. Then in the case where $\pi_2(M)=0$, for example, we can see that $\pi_1(\Omega_f(M))$ is isomorphic to the subgroup $\{a\in \pi_1(M) \mid f_*(a)=a\}$.

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    $\begingroup$ In the homotopy pullback you suggest you have $\Omega_f(M)\simeq M$ and there isn't much interesting to say about $\pi_1$. I think you need a different construction to capture the problem. $\endgroup$
    – Tyrone
    Feb 26 at 14:29
  • $\begingroup$ thanks a lot Mark. I'm wondering, is there a way to see from your second exact sequence that in the simply connected case, $\pi_1(\Omega_f(M))\cong \pi_2(M)$? $\endgroup$
    – Luigi M
    Feb 26 at 16:51
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    $\begingroup$ You're welcome, I wish I had more time to think about this interesting question. In the simply connected case my 2nd exact sequence shows that $\pi_1(\Omega_f(M))$ is isomorphic to the cokernel of a map $\pi_2(M\times M)\to \pi_2(M\times M)$, which I believe should be given by $(x,y)\mapsto(x-y,f_*(x)-y)$. $\endgroup$
    – Mark Grant
    Feb 26 at 21:41
  • $\begingroup$ no worries, I'm learning a lot from your question! thanks again $\endgroup$
    – Luigi M
    Feb 27 at 14:49

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