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).

  • 1
    $\begingroup$ Maybe this article could help. Notice that your space $X$ is a $\mathbb{Z} / 2 \mathbb{Z}$-space under the conjugation. $\endgroup$
    – Zerox
    Commented Apr 10, 2023 at 12:45

2 Answers 2


$\newcommand{\Z}{\mathbf{Z}}\newcommand{\Map}{\mathrm{Map}}$Let $\sigma$ denote the sign representation of $\Z/2$, and let $S^{d\sigma}$ denote the one-point compactification of $\sigma^{\oplus d}$. Let $X$ be a space with a $\Z/2$-action. It seems that you're interested in $(\Omega^{d\sigma} X)^{\Z/2}$ (this is your $S_d$). If $[d-1] = \{0,\cdots,d-1\}$, this can be described as the total homotopy fiber of the functor $2^{[d-1]} \to \mathrm{Spaces}_\ast$ sending $J \mapsto \Map_{\Z/2}((\Z/2)^J_+, X)$. This comes from the $\Z/2$-equivariant cell structure of $S^{d\sigma}$: it is the total homotopy cofiber of the functor $(2^{[d-1]})^\mathrm{op} \to \mathrm{Spaces}_\ast$ sending $J\mapsto (\Z/2)^J_+$.

For instance, when $d = 1$ (you can induct on $d$ to get the general claim from this case), recall that $S^\sigma$ can be constructed as the homotopy cofiber of the crushing map $(\Z/2)_+ \to S^0$. This implies that there's a fiber sequence $$(\Omega^\sigma X)^{\Z/2} = \Map_{\Z/2}(S^\sigma, X) \to \Map_{\Z/2}(S^0, X) = X^{\Z/2} \to \Map_{\Z/2}((\Z/2)_+, X) = X,$$ the second map being the inclusion of the fixed points. This is precisely what it means for $(\Omega^\sigma X)^{\Z/2}$ to be the total homotopy fiber of the functor $2^{[0]} = \Delta^1 \to \mathrm{Spaces}_\ast$. Note that this fiber sequence gives a long exact sequence on homotopy groups $$\cdots \to \pi_{j+1}(X) \to \pi_j((\Omega^\sigma X)^{\Z/2}) \to \pi_j(X^{\Z/2}) \to \pi_j(X) \to \cdots$$

For example, if $X = \mathrm{BU}(n)$ equipped with complex conjugation, this is going to imply that $(\Omega^\sigma \mathrm{BU}(n))^{\Z/2}$ is the homotopy fiber of the map $\mathrm{BO}(n) \to \mathrm{BU}(n)$, i.e., is $\mathrm{U}(n)/\mathrm{O}(n)$. In your case, if one equips $V_m(\mathbf{C}^n) = \mathrm{U}(n)/\mathrm{U}(n-m)$ with the $\Z/2$-action given by complex conjugation, this is going to imply that $(\Omega^\sigma V_m(\mathbf{C}^n))^{\Z/2}$ is the total homotopy fiber of the square $$\require{AMScd} \begin{CD} \mathrm{BO}(n-m) @>>> \mathrm{BU}(n-m)\\ @VVV @VVV \\ \mathrm{BO}(n) @>>> \mathrm{BU}(n). \end{CD}$$ Not sure if this has a simpler description.


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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