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Let $X$ be a topological space such that complex conjugation is defined (e.g. $\mathbb{C}^n$) and let us define the set of maps $$S_d:= \left\{f: (I^d,\partial I^d)\to (X,x_0)\mid \overline{f(k)} = f(-k)\right\} \\\subseteq \left\{ f: (I^d,\partial I^d)\to (X,x_0) \right\} = \Omega^dX,$$ where $I = [-1,1]$.

Equip the sets $S_d$ and $\Omega^dX$ with the Compact-open topology, such that they become topological spaces. What can we say about the homotopy groups $\pi_n(S_d,c_{x_0})$, where $c_{x_0}$ is the constant map into $x_0$?

I am looking for a strategy in computing $\pi_n(S_d,c_{x_0})$. What I do know are the homotopy groups $\pi_n(\Omega^dX,c_{x_0})\cong \pi_{n+d}(X,x_0)$, which is a standard result in homotopy theory. But $S_d$ is a subspace in $\Omega^dX$, which does not have to share the same homotopy groups. The elements of $S_d$ satisfy a certain $\mathbb{Z}_2$-equivariance condition and the theory about $G$-equivariant homotopy seems to be very involved, although I would certainly dive into it, when I knew that there were tools with which one could calculate the homotopy groups of $S_d$.

Thank you in advance!

Edit: Consider as an example $X = V_p(\mathbb{C}^q)$, the complex Stiefel manifold, whose elements we will interpret as complex $q\times p$-matrices.

Edit 2: The set of path components $\pi_0(S_d,c_{x_0})$ would be sufficient.

Edit 3: Question has been answered (see below).

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  • \( and \) were causing issues in your MathJax, now fixed (just use ( and )) – FShrike Apr 09 '23 at 15:30
  • Good luck with this anyway. I've found algebraic topology isn't always a well-received topic on MSE, and this is suitably advanced that you might get a (better and faster) answer on MathOverflow – FShrike Apr 09 '23 at 15:34
  • @FShrike thank you for your wishes and your advice but I think there are far more advanced topics discussed on MSE than this one, also in the rubric of algebraic topology. – Mathematics enthusiast Apr 09 '23 at 20:51
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    What precisely is "a topological space such that complex conjugation is defined" supposed to mean? Do you want to consider any space with an involution or do you have something more specific in mind? I'm also confused by your definition of $S_d$, since $-k$ is not in $I^d$ if $k\neq0$. Perhaps the maps should be defined on $[-1,1]^d$? – Thorgott Apr 22 '23 at 16:01
  • @Thorgott A general space with an involution (i.e. a $\mathbb{Z}_2$-action) would work, and something more specific is also already given in the first Edit: $V_p(\mathbb{C}^q)$, i.e. complex rectangular matrices with orthonormal columns. We definitively have $-k\in I^d$ because I have defined $I=[-1,1]$. Has this resolved your confusion? – Mathematics enthusiast Apr 22 '23 at 16:56
  • I completely missed the part where you clarified $I=[-1,1]$. Sorry. – Thorgott Apr 22 '23 at 17:29

1 Answers1

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My question has been answered in Lemma 3.1 in https://arxiv.org/abs/2404.09023

Indeed, $\pi_{D}\left(\left(\Omega^{d+1}X\right)^{\mathbb{Z}_2},c_{x_0}\right)\cong\pi_{D+1}\left(\Omega^{d}X,\left(\Omega^{d}X\right)^{\mathbb{Z}_2},c_{x_0}\right)$ for any $\mathbb{Z}_2$-space $X$ with $\mathbb{Z}_2$-fixed point $x_0\in X^{\mathbb{Z}_2}$ which gives rise to the long exact sequence of the pair $\left(\Omega^{d}X,\left(\Omega^{d}X\right)^{\mathbb{Z}_2}\right)$. The map $c_{x_0}$ denotes the constant map to $x_0$.

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