33
$\begingroup$

I want to prove the Hausdorff property of the projective space with this definition: the sphere $S^n$ with the antipodal points identified. It's seems easy, but I can't prove formally with this definition. I take neighborhoods of the points $p$, $q$ in $S^n/\sim $, with $p\neq q$. In $S^n$ we will have four points $p$, $q$, $-p$, $-q$, with 4 disjoints neighborhoods by the Hausdorff property of $\mathbb R^{n+1}$. I don't know how to prove with rigour why when we pass to the quotient we will have two disjoints neighborhoods of $p$ and $q$.

Thanks

$\endgroup$

9 Answers 9

35
$\begingroup$

Let $q:S^n\to P^n$ be the quotient map, and let $u,v\in P^n$ with $u\ne v$; there are $x,y\in S^n$ such that $q^{-1}[\{u\}]=\{x,-x\}$ and $q^{-1}[\{v\}]=\{y,-y\}$. Let $\epsilon=\frac13\min\{\|x-y\|,\|x+y\|\}$, and set

$$U=B(x,\epsilon)\cap S^n\quad\text{and}\quad V=B(y,\epsilon)\cap S^n\;,$$

where the open balls are taken in $\Bbb R^n$. Then $U,V,-U$, and $-V$ are pairwise disjoint open nbhds of $x,y,-x$, and $-y$, respectively, in $S^n$. Moreover, $q^{-1}\big[q[U]\big]=-U\cup U$ and $q^{-1}\big[q[V]\big]=-V\cup V$. Show that $q[U]$ and $q[V]$ are disjoint open nbhds of $u$ and $v$ in $P^n$.

$\endgroup$
13
  • $\begingroup$ Why did you take such a $\epsilon$? $\endgroup$
    – user42912
    Nov 22, 2012 at 0:46
  • $\begingroup$ @Rafael: It was convenient. I could have used any $\epsilon\le\frac12\min\{\|x-y\|,\|x+y\|\}$. $\endgroup$ Nov 22, 2012 at 0:50
  • $\begingroup$ yes, of course, thank you :) $\endgroup$
    – user42912
    Nov 22, 2012 at 0:56
  • 1
    $\begingroup$ @BrianM.Scott In particular, math.stackexchange.com/questions/1699700/… $\endgroup$ Mar 16, 2016 at 6:06
  • 1
    $\begingroup$ @Anthony: Yes, that should work. My own inclination would be simply to normalize $x$ and $y$, at which point you have essentially the situation with which I was working here, but that is perhaps a matter of taste. $\endgroup$ Mar 16, 2016 at 6:25
24
$\begingroup$

Let $G$ be a compact Hausdorff group. Let $X$ be a locally compact Hausdorff space. Suppose $G$ acts continuously on $X$. We will prove that the orbit space $X/G$ is Hausdorff. The proof is due to Bourbaki. As an immediate corollary, projective spaces over $\mathbb{R}$ and $\mathbb{C}$ are Hausdorff.

Definition 1 Let $X, Y$ be locally compact Haudsorff spaces. Let $f\colon X \rightarrow Y$ be a continuous map. If $f^{-1}(K)$ is compact for every compact subset $K$ of $Y$, $f$ is called a proper map.

Lemma 1 Let $X$ be a topological space. Let $(A_i)_{i\in I}$ be a family of subsets of $X$. Suppose $X = \bigcup_i int(A_i)$, where $int(A_i)$ is the interior of $A_i$. Let $B$ be a subset of $X$. Suppose $B \cap A_i$ is closed in $A_i$ for every $i \in I$. Then $B$ is closed in $X$.

Proof: Let $C = X - B$. Since $C \cap A_i = A_i - (B \cap A_i)$, $C \cap A_i$ is open in $A_i$. Hence there exists an open subset $V_i$ of $X$ such that $C \cap A_i = V_i \cap A_i$. Then $C \cap int(A_i) = (C \cap A_i) \cap int(A_i) = (V_i \cap A_i) \cap int(A_i) = V_i \cap int(A_i)$. Hence $C \cap int(A_i)$ is open in $X$. Since $X = \bigcup int(A_i)$, $C = \bigcup (C \cap int(A_i))$. Hence $C$ is open in $X$. Hence $B$ is closed in $X$. QED

Lemma 2 Let $X, Y$ be topological spaces. Let $f\colon X \rightarrow Y$ be a continuous map. Let $(A_i)_{i\in I}$ be a family of subsets of $Y$. Suppose $Y = \bigcup_i int(A_i)$. Let $f_i\colon f^{-1}(A_i) \rightarrow A_i$ be the restriction of $f$ for each $i \in I$. Suppose $f_i$ is a closed map for every $i \in I$. Then $f$ is a closed map.

Proof: Let $B$ be a closed subset of $X$. Since $f(B) \cap A_i = f_i(B \cap f^{-1}(A_i))$, $f(B) \cap A_i$ is closed in $A_i$ for every $i \in I$. Hence $f$ is a closed map by Lemma 1. QED

Lemma 3 Let $X, Y$ be locally compact Hausdorff spaces. Let $f\colon X \rightarrow Y$ be a proper map. Then $f$ is a closed map.

Proof: There exists a family $(A_i)_{i\in I}$ of compact subsets of $Y$ such that $Y = \bigcup_i int(A_i)$.Let $f_i\colon f^{-1}(A_i) \rightarrow A_i$ be the restriction of $f$ for each $i \in I$. Since $f$ is proper, $f^{-1}(A_i)$ is compact. Hence $f_i$ is a closed map. Hence $f$ is a closed map by Lemma 2. QED

Definition 2 Let $G$ be a locally compact Hausdorff group. Let $X$ a locally compact Hausdorff space. Suppose $G$ acts on $X$ continuously. Suppose the map $f\colon G\times X \rightarrow X\times X$ defined by $f(s, x) = (x, sx)$ is proper. Then we say $G$ acts properly on $X$.

Definition 3 Suppose a group $G$ acts on a set $X$. Let $K, L$ be subsets of $X$. We denote $P(K, L) = \{s \in G| sK \cap L \neq \emptyset\}$.

Lemma 4 Let $X, Y$ be Hausdorff spaces. Let $C$ be a compact subset of $X\times Y$. Then there exists a compact subset $K$(resp. $L$) of $X$(resp. $L$) such that $C \subset K\times L$.

Proof: Let $K$(resp. $L$) be the image of $C$ by the projection map on the first(resp. the second) factor of $X\times X$. Then $K$ and $L$ are compact and $C \subset K\times L$. QED

Lemma 5 Let $f\colon X_1 \rightarrow Y_1$(resp. $g\colon X_2 \rightarrow Y_2$) be a proper map. Then $f\times g\colon X_1\times X_2 \rightarrow Y_1 \times Y_2$ is proper.

Proof: Let $C$ be a compact subset of $Y_1\times Y_2$. By Lemma 4, there exists a compact subset $K$(resp. $L$) of $Y_1$(resp. $Y_2$) such that $C \subset K\times L$. Since $(f\times g)^{-1}(K\times L) = f^{-1}(K)\times g^{-1}(L)$ is compact, $(f\times g)^{-1}(C)$ is compact. QED

Lemma 6 Let $X$ be a compact Haudsorff space. Let $Y$ be a locally compact Hausdorff space. Then the projection map $\pi\colon X\times Y \rightarrow Y$ is proper.

Proof: Let $p$ be a one point space. Then $f\colon X \rightarrow p$ is proper. Hence, by Lemma 5, $\pi = f\times id_Y\colon X \times Y \rightarrow p\times Y = Y$ is proper. QED

Lemma 7 Let $G$ be a locally compact Hausdorff group. Let $X$ a locally compact Hausdorff space. Suppose $G$ acts on $X$ continuously. Then $P(K, L)$ is closed for any compact subsets $K, L$ of $X$.

Proof: Let $f\colon G\times X \rightarrow X$ be the map defined by $f(s, x) = sx$. Since $f$ is continuous, $f^{-1}(L)$ is closed. Let $\pi\colon G\times K \rightarrow G$ be the projection. By Lemma 6, $\pi$ is proper. Hence, by Lemma 3, $\pi$ is a closed map. Hence $P(K, L) = \pi(f^{-1}(L))$ is closed. QED

Lemma 8 Let $G$ be a locally compact Hausdorff group. Let $X$ a locally compact Hausdorff space. Suppose $G$ acts on $X$ continuously. Suppose $P(K, L)$ is compact for any compact subsets $K, L$ of $X$. Then $G$ acts properly on $X$.

Proof: Let $h\colon G\times X \rightarrow X\times X$ be the map defined by $h(s, x) = (x, sx)$. Let $C$ be a compact subset of $X\times X$. By Lemma 4, there exist compact subsets $K, L$ of $X$ such that $C \subset K\times L$. Then $h^{-1}(K\times L) \subset P(K, L)\times K$. Since $P(K, L)\times K$ is compact and $h^{-1}(K\times L)$ is closed, $h^{-1}(K\times L)$ is compact. Since $C \subset K\times L$, $h^{-1}(C) \subset h^{-1}(K\times L)$. Since $h^{-1}(C)$ is closed, it is compact. Hence $h$ is proper. Hence $G$ acts proplerly on $X$. QED

Lemma 9 Let $X$ be a topological space. Let $R$ be an equivalence relation on $X$. Let $X/R$ be the quotient space. Let $\pi\colon X \rightarrow X/R$ be the canonical map. Let $\Gamma = \{(x, y)\in X\times X| x \equiv y$ (mod $R)\}$. Suppose $\pi$ is an open map and $\Gamma$ is closed. Then $X/R$ is Hausdorff.

Proof: Let $\rho := \pi\times \pi\colon X\times X \rightarrow (X/R)\times (X/R)$. Since $\pi$ is an open map, so is $\rho$. Since $\rho(\Gamma)$ is the diagonal subset of $(X/R)\times (X/R)$, it suffices to prove that $\rho(\Gamma)$ is closed. Since $\Gamma = \rho^{-1}(\rho(\Gamma)$, $X\times X - \Gamma = \rho^{-1}((X/R)\times (X/R) - \rho(\Gamma))$. Since $\rho$ is surjective, $\rho(X\times X - \Gamma) = (X/R)\times (X/R) - \rho(\Gamma)$. Since $\rho$ is an open map and $X\times X - \Gamma$ is open, $(X/R)\times (X/R) - \rho(\Gamma)$ is open. Hence $\rho(\Gamma)$ is closed. QED

Lemma 10 Let $G$ be a locally compact Hausdorff group. Let $X$ a locally compact Hausdorff space. Suppose $G$ acts properly on $X$. Then the orbit space $X/G$ is Hausdorff.

Proof: Let $\pi\colon X \rightarrow X/G$ be the canonical map. Let $U$ be an open subset of $X$. Then $\pi^{-1}(\pi(U)) = GU$. Since $GU$ is open, $\pi$ is an open map. By Lemma 9, it suffices to prove that $\Gamma = \{(x, y) \in X\times X| x \equiv y$ (mod $G)\}$ is closed. Let $h\colon G\times X \rightarrow X\times X$ be the map defined by $h(s, x) = (x, sx)$. Since $G$ acts properly on $X$, $h$ is proper. Hence $h$ is closed by Lemma 3. Hence $\Gamma = h(G\times X)$ is closed. QED

Proposition Let $G$ be a compact Hausdorff group. Let $X$ be a locally compact Hausdorff space. Suppose $G$ acts continuously on $X$. Then $G$ acts properly on $X$. Hence the orbit space $X/G$ is Hausdorff.

Proof: By Lemma 7, $P(K, L)$ is closed for any compact subsets $K, L$ of $X$. Since $G$ is compact, $P(K, L)$ is compact. Hence, by Lemma 8, $G$ acts proplerly on $X$. Hence, by Lemma 10, $X/G$ is Hausdorff. QED

Corollary Let $K$ be the field of real numbers or the field of complex numbers. Then the projective space $P^n(K)$ over $K$ is Hausdorff.

Proof: Let $x = (x_1, \dots, x_n) \in K^n$. We denote $||x|| = (\sum_i x_i \bar x_i)^{\frac{1}{2}}$. Let $G = \{x \in K|\ |x| = 1\}$. Then $G$ is a subgroup of the multiplicative group of $K$. Let $X = \{x \in K^n|\ ||x|| = 1\}$. Clearly $G$ acts on $X$ continuously. Since $G$ and $X$ are compact Hausdorff spaces, $X/G$ is Hausdorff by the proposition. Since $P^n(K) = X/G$, the assertion follows. QED

Remark 1 The above proposition holds without assuming $X$ is locally Hausdorff, but only assuming it is Hausdorff as the answer to this question shows.

Remark 2 If $G$ and $X$ are both compact Hausdorff, it is easy to prove that $X/G$ is Hausdorff. Let $\Gamma = \{(x, y) \in X\times X| x \equiv y$ (mod $G)\}$. Let $h\colon G\times X \rightarrow X\times X$ be the map defined by $h(s, x) = (x, sx)$. Since $h$ is continuous, $\Gamma = h(G\times X)$ is compact, hence closed. By Lemma 9, $X/G$ is Hausdorff.

$\endgroup$
6
  • 3
    $\begingroup$ The theory of proper actions is essentially due to Palais, On the existence of slices for actions of non-compact Lie groups. Another nice exposition is in Koszul's TIFR Lectures on Groups of transformations. $\endgroup$
    – commenter
    Nov 21, 2012 at 13:16
  • $\begingroup$ @commenter Thanks for the nice links. $\endgroup$ Nov 22, 2012 at 6:52
  • $\begingroup$ Great answer, I was interested in the same and found this using google. Thanks Makoto Kato!!! $\endgroup$ Dec 12, 2013 at 11:05
  • $\begingroup$ @FabianWerner You're welcome. :) $\endgroup$ Dec 12, 2013 at 18:40
  • $\begingroup$ In the corollary, what do you mean by $|x|=1$ in the definition of $G$, is that norm? $\endgroup$
    – user135520
    Feb 14, 2016 at 15:32
8
$\begingroup$

You did ask explicitly for a proof that uses the construction of projective space as a quotient of $S^n$, but consider this for fun. If you are willing to consider the projective space as the set of lines through $0$ in $\mathbb R^{n+1}$, then there is an intuative geometric way of seeing the Hausdorff property, that one should be able to convert into a rigorous proof:

Let $x$ and $y$ be distinct points in $\mathbb P^n$. Let $l_x$ and $l_y$ be the corresponding lines in $\mathbb R^{n+1}$. The Hausdorff property of projective space follows from the fact that we can fit the lines into two cones in $\mathbb R^{n+1}$ that only have $0$ in common, whose projections to projective space then give disjoint open sets that contain $x$ and $y$. It's helpful to draw a picture of the 2-dimensional situation to get a feel for this; to construct the cones you can fix an inner product and take the set of lines whose angle with $l_x$ (resp. $l_y$) is less than some $\epsilon > 0$. (By doing this we're basically making $\mathbb P^n$ into a metric space, whence the Hausdorff property follows.)

$\endgroup$
2
  • $\begingroup$ Why cones? the neighborhood of the points of $\mathbb P^n$ in this case are cones? thanks for your answer $\endgroup$
    – user42912
    Nov 22, 2012 at 0:51
  • $\begingroup$ I picked cones for ease of use, but you can use whatever disjoint open sets that contain each line you want. The idea is just to get open sets on $\mathbb P^n$ by building them "upstairs" in the quotient topology. It really helps to draw a picture in $\mathbb R^2$ to see what's going on! $\endgroup$ Nov 22, 2012 at 8:44
7
$\begingroup$

Recall that an equivalence relation $\sim$ on a topological space X is said to be $\textit{open}$ if for every open subset $A$ of $X$, the set $$[A] := \{x \in X | x \sim a \text{ for some a} \in A\}$$ is also open. It is easy to see that group actions on topological spaces give open equivalence relations. It is not hard to show the following

$\textbf{Lemma. }$ Let $\sim$ be an open equivalence relation on a space $X$. Then $X /\sim$ is Hausdorff iff the set $\{(x,y)| x \sim y\}$ is closed in $X \times X$.

Let $X = \mathbb{R}^{n+1} - 0$ and $\sim$ be the quotient by the action of $\mathbb{R}^{+}$ by scaling. Define a real valued function $f$ on $X \times X$ as $$f(x_1, ..., x_{n+1}, y_1,...,y_{n+1}) = \sum_{i \neq j}(x_i y_j - y_i x_j)^2 . $$Then $f(x,y)$ is clearly continuous and vanishes iff $y = \lambda x$ for some $\lambda \in \mathbb{R}^{+}$, that is, iff $x \sim y.$ Thus we have that $$f^{-1}(0) = \{(x,y)| x \sim y\} $$ is closed in $X \times X$ and by the lemma, $\mathbb{RP}^n$ is Hausdorff.

$\endgroup$
1
  • $\begingroup$ Instead of R^+ positive reals under usual multiplication, it should be R* nonzero reals with under usual multiplication $\endgroup$
    – Swapnil
    Sep 18, 2022 at 11:22
5
$\begingroup$

Another way to see this is to note that $\mathbb{R}P^n$ is a CW-complex with one cell in each dimension $i\leq n$ i.e

$$\mathbb{R}P^n=\bigcup_{i=0} ^n e^n.$$

Since CW-complexes are Hausdorff, $\mathbb{R}P^n$ is.

$\endgroup$
3
  • $\begingroup$ where can I see this definition of $\mathbb RP^n$? thanks for your answer $\endgroup$
    – user42912
    Nov 22, 2012 at 0:54
  • $\begingroup$ @rafaelchavez on page 6 of: math.cornell.edu/~hatcher/AT/AT.pdf $\endgroup$ Nov 22, 2012 at 8:15
  • $\begingroup$ +1 for pointing out this wonderful construction! This is the shortest and (in my opinion) the most elegant proof of the Hausdorff-ness of $\Bbb RP^n$. $\endgroup$ Jul 18, 2023 at 12:14
5
$\begingroup$

Perhaps the nicest approach to prove that $\Bbb{R}P^n$ is Hausdorff is to notice that it is the orbit space $S^n/(\Bbb{Z}/2\Bbb{Z})$ where the action of $\Bbb{Z}/2\Bbb{Z}$ is given by the antipodal map. The following problem (which you should try to do) will now complete the proof that real projective space is Hausdorff:

Let $G$ be a finite group that acts on a Hausdorff space $X$. Then the orbit space $X/G$ is Hausdorff.

$\endgroup$
2
$\begingroup$

Let $K$ be the field of real numbers or the field of complex numbers. Let $G = K^*$ be the multiplicative group of $K$. Let $X = K^{n+1} - {0}$. Then $G$ acts on $X$. We regard $X$ as a topological subspace of $K^{n+1}$. Then the projective space $P^n$ over $K$ is, by defintion, the orbit space $X/G$ with the quotient topology.

Let $\pi\colon X \rightarrow P^n$ be the canonical map. Let $x = (x_0,x_1,\dots,x_n) \in X$. We denote $\pi(x)$ by $[x_0,x_1,\dots,x_n]$. Let $\Delta = \{(x,x) \in P^n\times P^n| x \in P^n\}$. To prove that $P^n$ is Hausdorff, it suffices to prove that $\Delta$ is closed in $P^n\times P^n$.

Let $([x_0,x_1,\dots,x_n], [y_0,y_1,\dots,y_n]) \in \Delta$. Suppose $x_i \neq 0$. Then $y_i \neq 0$ and $x_j/x_i = y_j/y_i$ for $i \neq j$. Hence $x_iy_j = x_jy_i$ for $i \neq j, 0\le i, j\le n$.

Conversely suppose $([x_0,x_1,\dots,x_n], [y_0,y_1,\dots,y_n]) \in P^n\times P^n$. Suppose $x_iy_j = x_jy_i$ for $i \neq j, 0\le i, j\le n$. Suppose $x_i \neq 0$. If $y_i = 0$, then $y_j = 0$ for all $j$. This is a contradiction. Hence $y_i \neq 0$. Hence $x_j/x_i = y_j/y_i$ for all $j \neq i$. Hence $([x_0,x_1,\dots,x_n], [y_0,y_1,\dots,y_n]) \in \Delta$.

Therefore $\Delta = \{([x_0,x_1,\dots,x_n], [y_0,y_1,\dots,y_n]) \in P^n\times P^n| x_iy_j = x_jy_i, i\neq j, 0\le i,j\le n\}$. Let $U_i = \{[x_0,x_1,\dots,x_n] \in P^n|\ x_i \neq 0\}$. Then $U_i$ is an open subset of $P^n$ and it is is canonically isomorphic to $K^n$(see here). Clearly $(U_i)_{i \in I}$ is an open cover of $P^n$, where $I = \{0,1,\dots,n\}$. Hence $(U_i\times U_j)_{(i, j) \in I\times I}$ is an open cover of $P^n\times P^n$. Since $\Delta$ is the common zeros of bihomogeneous polynomials $x_iy_j - x_jy_i, i\neq j$, $(U_i\times U_j) \cap \Delta$ is canonically isomorphic to an algebraic subset of $K^n\times K^n$(e.g. see Mychael Joyce's answer to this question). Hence it is closed in $U_i\times U_j$. Hence $\Delta$ is closed.

$\endgroup$
1
$\begingroup$

The proof I'll give bellow works for the fields $\mathbb K = \mathbb R$ and $\mathbb C$. It also works for $\mathbb K = \mathbb H$. Let $S$ be the unit sphere in $\mathbb K^{n+1}$ and $S_{\mathbb K}$ be the unit sphere in $\mathbb K$, i.e., $$ S_{\mathbb R} = \{-1, 1\}\quad \text{and}\quad \quad S_{\mathbb C} = \mathbb S^1.$$

We have the surjective map $\pi:S \to \mathbb P_{\mathbb K}^n$ given by $\pi(x) = [x]$. The topology considered in the projective space is the quotient topology. Notice that if $U \subset S$ is open, then $$\pi^{-1}(\pi(U)) = \cup_{\gamma\in S_{\mathbb K}} \gamma U$$ is open and therefore $\pi$ is an open map.

Now, take $[x],[y] \in \mathbb P_{\mathbb K}^n$, $[x] \neq [y]$. Notice $$\pi^{-1}[x] = \{\gamma x \in S:\gamma \in S_{\mathbb K}\} = S_{\mathbb K} x.$$

The map $S_{\mathbb K} \to \pi^{-1}[x]$ given by $\gamma \mapsto \gamma x$ is homeomorphism and therefore $\pi^{-1}[x]$ is compact. So, we have $\pi^{-1}[x]$ and $\pi^{-1}[y]$ are closed subsets of $S$. Clearly they are also disjoint.

By the Urysohn's lemma, there is a continuous function $g:S \to \mathbb R$ such that $g|_{\pi^{-1}[x]} = 1$ and $g|_{\pi^{-1}[y]} = 0$.

But this function is not necessarly invariant under action of $S_{\mathbb K}$, so we define $h: S \to \mathbb R$ by the formula $$h(x) = \int_{S_{\mathbb K}} g(\gamma x) d\gamma,$$ where $d\gamma$ is the Haar measure of $S_{\mathbb K}$. More explicitly, $$h(x) = \frac{1}{2}(g(x)+g(-x))\quad \text{for} \quad \mathbb K = \mathbb R$$ and $$h(x) = \frac{1}{2\pi}\int_{0}^{2\pi} g(e^{i\theta}x)d\theta\quad \text{for} \quad \mathbb K = \mathbb C.$$

Now, since the sets $\pi^{-1}[x]$ and $\pi^{-1}[y]$ are invariant under action of $S_{\mathbb K}$, we have $h|_{\pi^{-1}[x]} = 1$ and $h|_{\pi^{-1}[y]} = 0$.

This function $h$ is a function on the projective space. More precisely, define $\widehat h: \mathbb P_{\mathbb K}^n \to \mathbb R$ by the formula $\widehat h([z]) = h(z)$. Since, $h(\gamma z) = h(z)$ for $\gamma \in S_{\mathbb K}$, the function $\widehat h$ is well defined. Besides that, since $h = \widehat h \circ \pi$ and $\pi$ is open map, we obtain $\widehat h$ is continuous.

Taking the open sets $U = \widehat h^{-1}(1/2,\infty)$ and $V = \widehat h^{-1}(-\infty,1/2)$ we have $[x] \in U$, $[y] \in V$ and $U \cap V = \varnothing$, proving $\mathbb P_{\mathbb K}^n$ is Hausdorff.

$\endgroup$
0
$\begingroup$

I think in this case, and in may others, it is way easier to use the following equaivalent definition of Hasudorff:

Definition: A topological space X is Hausdorff iff the Diagonal $\,\Delta_X:=\{(x,x)\in X^2\}\,$ is closed in the product space $\,X\times X\,$

$\endgroup$
2
  • $\begingroup$ How does this help here? $\endgroup$
    – WillG
    May 19, 2021 at 6:36
  • $\begingroup$ @WillG Decisively...but I can hardly remember that. $\endgroup$
    – DonAntonio
    May 19, 2021 at 12:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .