[proofplan] We verify directly that the standard affine coordinate charts cover [projective space](/page/Projective%20Space) and that their coordinate changes are regular on their overlap domains. Each chart is inverted by inserting a $1$ in the normalized coordinate slot, so its image is all of $K^n$. On an overlap, the transition map divides every normalized coordinate by the coordinate corresponding to the target chart index; the overlap condition is exactly that this denominator is nonzero. These transition maps are smooth when $K=\mathbb{R}$ and holomorphic when $K=\mathbb{C}$, giving the claimed manifold structures and dimensions. [/proofplan]

[step:Identify each affine chart with $K^n$]Fix an integer $i$ with $0 \leq i \leq n$. For a point $[x_0:\cdots:x_n] \in U_i$, define $\varphi_i([x_0:\cdots:x_n])$ to be the element of $K^n$ obtained from \begin{align*} \left(\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}\right) \end{align*} by deleting the $i$-th coordinate. This map is well-defined. Indeed, if $[y_0:\cdots:y_n]=[x_0:\cdots:x_n]$, then there exists $\lambda \in K^\times$ such that $y_m=\lambda x_m$ for every $0 \leq m \leq n$. Since $x_i \neq 0$, also $y_i=\lambda x_i \neq 0$, and for every $m$, \begin{align*} \frac{y_m}{y_i}=\frac{\lambda x_m}{\lambda x_i}=\frac{x_m}{x_i}. \end{align*} Thus the normalized coordinates do not depend on the representative. Define the inverse candidate $\psi_i: K^n \to U_i$ as follows. For $a=(a_0,\ldots,a_{n-1}) \in K^n$, let $\widetilde a \in K^{n+1}$ be the vector whose $i$-th coordinate is $1$, whose coordinates before slot $i$ are $a_0,\ldots,a_{i-1}$, and whose coordinates after slot $i$ are $a_i,\ldots,a_{n-1}$. Then set \begin{align*} \psi_i(a):=[\widetilde a]. \end{align*} The $i$-th coordinate of $\widetilde a$ is $1$, so $\psi_i(a)\in U_i$. Direct substitution gives $\varphi_i(\psi_i(a))=a$ for every $a\in K^n$. Conversely, if $[x_0:\cdots:x_n]\in U_i$, then $\psi_i(\varphi_i([x_0:\cdots:x_n]))$ is represented by \begin{align*} \left(\frac{x_0}{x_i},\ldots,\frac{x_{i-1}}{x_i},1,\frac{x_{i+1}}{x_i},\ldots,\frac{x_n}{x_i}\right), \end{align*} which represents the same projective point as $(x_0,\ldots,x_n)$ because it is $x_i^{-1}(x_0,\ldots,x_n)$. Hence $\psi_i=\varphi_i^{-1}$, and $\varphi_i$ is a bijection from $U_i$ onto $K^n$.[/step]

[guided]The purpose of this step is to check that the proposed projective coordinates really are coordinates. Fix $i$ with $0 \leq i \leq n$. A point of $U_i$ has a representative $(x_0,\ldots,x_n)\in K^{n+1}\setminus\{0\}$ with $x_i\neq 0$, so we may normalize by dividing all coordinates by $x_i$. This produces \begin{align*} \left(\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}\right), \end{align*} whose $i$-th coordinate is $1$. Deleting that coordinate leaves an element of $K^n$, and this is $\varphi_i([x_0:\cdots:x_n])$. We must first verify that this does not depend on the representative. Suppose $(y_0,\ldots,y_n)$ is another representative of the same projective point. Then there is $\lambda\in K^\times$ such that $y_m=\lambda x_m$ for every $m$. Since $x_i\neq 0$, also $y_i\neq 0$, and for each coordinate index $m$ we have \begin{align*} \frac{y_m}{y_i}=\frac{\lambda x_m}{\lambda x_i}=\frac{x_m}{x_i}. \end{align*} Therefore the normalized coordinate tuple is independent of the chosen representative. Now define an inverse map $\psi_i:K^n\to U_i$. Given $a=(a_0,\ldots,a_{n-1})\in K^n$, insert a $1$ in the $i$-th coordinate slot to obtain a vector $\widetilde a\in K^{n+1}$. Explicitly, the coordinates before slot $i$ are $a_0,\ldots,a_{i-1}$, the $i$-th coordinate is $1$, and the coordinates after slot $i$ are $a_i,\ldots,a_{n-1}$. Set \begin{align*} \psi_i(a):=[\widetilde a]. \end{align*} Because the $i$-th coordinate of $\widetilde a$ is $1$, this projective point lies in $U_i$. Applying $\varphi_i$ to $\psi_i(a)$ normalizes by the already normalized coordinate $1$ and then deletes the $i$-th coordinate, so $\varphi_i(\psi_i(a))=a$. Conversely, if $p=[x_0:\cdots:x_n]\in U_i$, then $\psi_i(\varphi_i(p))$ is represented by the normalized vector \begin{align*} \left(\frac{x_0}{x_i},\ldots,\frac{x_{i-1}}{x_i},1,\frac{x_{i+1}}{x_i},\ldots,\frac{x_n}{x_i}\right). \end{align*} This vector is $x_i^{-1}(x_0,\ldots,x_n)$, so it represents the same projective point as $(x_0,\ldots,x_n)$. Thus $\psi_i(\varphi_i(p))=p$. The maps $\varphi_i$ and $\psi_i$ are inverse bijections, so $U_i$ is identified with $K^n$.[/guided]

[step:Show that the standard charts cover projective space] Let $[x_0:\cdots:x_n]\in \mathbb{P}^n_K$. By definition of projective space, the representative $(x_0,\ldots,x_n)$ is not the zero vector. Hence there exists an index $i$ with $0\leq i\leq n$ such that $x_i\neq 0$. Therefore $[x_0:\cdots:x_n]\in U_i$, and \begin{align*} \mathbb{P}^n_K=U_0\cup\cdots\cup U_n. \end{align*} [/step]

[step:Compute the transition map on each nonempty overlap]Fix indices $i$ and $j$ with $0\leq i,j\leq n$. Let \begin{align*} V_{ij}:=\varphi_i(U_i\cap U_j)\subset K^n. \end{align*} For $a=(a_0,\ldots,a_{n-1})\in K^n$, let $\widetilde a\in K^{n+1}$ denote the vector obtained by inserting $1$ in the $i$-th coordinate slot, as in the inverse map $\psi_i=\varphi_i^{-1}$. Then $a\in V_{ij}$ exactly when the $j$-th coordinate of $\widetilde a$ is nonzero. Define the coordinate-reading function $\delta_{ij}:K^n\to K$ by \begin{align*} \delta_{ij}(a):=\widetilde a_j, \end{align*} where $\widetilde a_j$ is the $j$-th coordinate of $\widetilde a$. Thus \begin{align*} V_{ij}=\{a\in K^n:\delta_{ij}(a)\neq 0\}. \end{align*} If $i=j$, then $\delta_{ii}(a)=1$ for all $a\in K^n$, so $V_{ii}=K^n$ and $\varphi_i\circ\varphi_i^{-1}$ is the identity map on $K^n$. Assume now that $i\neq j$. For $a\in V_{ij}$, the transition map \begin{align*} \tau_{ji}:V_{ij}\to V_{ji} \end{align*} is defined by \begin{align*} \tau_{ji}:=\varphi_j\circ\varphi_i^{-1}. \end{align*} Since $\varphi_i^{-1}(a)=[\widetilde a]$ and $\widetilde a_j=\delta_{ij}(a)\neq 0$, normalizing by the $j$-th coordinate gives \begin{align*} \left(\frac{\widetilde a_0}{\widetilde a_j},\ldots,\frac{\widetilde a_n}{\widetilde a_j}\right). \end{align*} The value $\tau_{ji}(a)$ is obtained from this tuple by deleting its $j$-th coordinate. Hence every component of $\tau_{ji}$ is either the constant function $1/\delta_{ij}$, one of the coordinate functions $a_m/\delta_{ij}$ after the appropriate index shift, or a coordinate function divided by $\delta_{ij}$. In all cases, each component is a rational function on $K^n$ whose denominator is $\delta_{ij}$.[/step]

[guided]The transition map answers the following question: if a projective point is written in the $i$-th normalized coordinates, what are its $j$-th normalized coordinates? Fix $i$ and $j$. Let \begin{align*} V_{ij}:=\varphi_i(U_i\cap U_j). \end{align*} This is the part of the $i$-th coordinate space where the same projective point also lies in the $j$-th chart. Given $a=(a_0,\ldots,a_{n-1})\in K^n$, the inverse chart $\varphi_i^{-1}$ inserts a $1$ into the $i$-th slot and produces a representative $\widetilde a\in K^{n+1}$. The point belongs to $U_j$ exactly when its $j$-th homogeneous coordinate is nonzero. Therefore, if we define \begin{align*} \delta_{ij}:K^n\to K \end{align*} by setting $\delta_{ij}(a)$ equal to the $j$-th coordinate of $\widetilde a$, then the overlap domain in $i$-coordinates is \begin{align*} V_{ij}=\{a\in K^n:\delta_{ij}(a)\neq 0\}. \end{align*} When $i=j$, there is no real coordinate change: the map is $\varphi_i\circ\varphi_i^{-1}$, which is the identity map on $K^n$. Now suppose $i\neq j$. For $a\in V_{ij}$, the projective point is represented in homogeneous coordinates by $\widetilde a$, and the condition $a\in V_{ij}$ says precisely that $\widetilde a_j\neq 0$. To pass to the $j$-th chart, we normalize by dividing every homogeneous coordinate by $\widetilde a_j=\delta_{ij}(a)$. This gives the normalized tuple \begin{align*} \left(\frac{\widetilde a_0}{\widetilde a_j},\ldots,\frac{\widetilde a_n}{\widetilde a_j}\right). \end{align*} The $j$-th entry of this tuple is $1$, so $\varphi_j$ deletes that entry. The remaining entries are the components of \begin{align*} \tau_{ji}:=\varphi_j\circ\varphi_i^{-1}:V_{ij}\to V_{ji}. \end{align*} Thus every component of $\tau_{ji}$ is obtained by taking a coordinate of $\widetilde a$ and dividing it by $\delta_{ij}(a)$. Since the coordinates of $\widetilde a$ are either coordinate functions on $K^n$ or the constant function $1$, every component of the transition map is a rational function with denominator $\delta_{ij}$. The denominator is nonzero exactly on $V_{ij}$, which is why the formula is valid on the whole overlap and nowhere outside it.[/guided]

[step:Verify smooth compatibility for $K=\mathbb{R}$] Assume $K=\mathbb{R}$. For each pair $i,j$, the overlap domain \begin{align*} V_{ij}=\{a\in \mathbb{R}^n:\delta_{ij}(a)\neq 0\} \end{align*} is open in $\mathbb{R}^n$, because $\delta_{ij}:\mathbb{R}^n\to\mathbb{R}$ is either a coordinate projection or the constant function $1$, and $\mathbb{R}\setminus\{0\}$ is open in $\mathbb{R}$. The transition map $\tau_{ji}:V_{ij}\to V_{ji}$ has components which are rational functions with nonvanishing denominator $\delta_{ij}$ on $V_{ij}$. Polynomial functions on $\mathbb{R}^n$ are smooth, the reciprocal map $t\mapsto t^{-1}$ is smooth on $\mathbb{R}\setminus\{0\}$, and smooth functions are closed under products and composition. Therefore every transition map $\tau_{ji}$ is smooth. Since the charts cover $\mathbb{P}^n_{\mathbb{R}}$, each chart is a bijection onto the model space $\mathbb{R}^n$, and all transition maps are smooth on open overlap domains, the standard projective atlas defines a smooth atlas modeled on $\mathbb{R}^n$. Hence $\mathbb{P}^n_{\mathbb{R}}$ carries the corresponding smooth manifold structure of real dimension $n$. [/step]

[step:Verify holomorphic compatibility for $K=\mathbb{C}$] Assume $K=\mathbb{C}$. For each pair $i,j$, the overlap domain \begin{align*} V_{ij}=\{a\in \mathbb{C}^n:\delta_{ij}(a)\neq 0\} \end{align*} is open in $\mathbb{C}^n$, because $\delta_{ij}:\mathbb{C}^n\to\mathbb{C}$ is either a coordinate projection or the constant function $1$, and $\mathbb{C}\setminus\{0\}$ is open in $\mathbb{C}$. The transition map $\tau_{ji}:V_{ij}\to V_{ji}$ has components which are rational functions with nonvanishing denominator $\delta_{ij}$ on $V_{ij}$. Polynomial functions on $\mathbb{C}^n$ are holomorphic, the reciprocal map $z\mapsto z^{-1}$ is holomorphic on $\mathbb{C}\setminus\{0\}$, and holomorphic functions are closed under products and composition. Therefore every transition map $\tau_{ji}$ is holomorphic. Since the charts cover $\mathbb{P}^n_{\mathbb{C}}$, each chart is a bijection onto the model space $\mathbb{C}^n$, and all transition maps are holomorphic on open overlap domains, the standard projective atlas defines a holomorphic atlas modeled on $\mathbb{C}^n$. Hence $\mathbb{P}^n_{\mathbb{C}}$ carries the corresponding complex manifold structure of complex dimension $n$. [/step]

[step:Handle the zero-dimensional case] When $n=0$, the projective space $\mathbb{P}^0_K$ has one homogeneous coordinate and every nonzero vector in $K^1$ represents the same point. Thus $\mathbb{P}^0_K$ is a singleton. There is one chart $U_0=\mathbb{P}^0_K$, and $\varphi_0$ maps it bijectively to $K^0$, the one-point [vector space](/page/Vector%20Space). There are no nontrivial transition maps, so the compatibility condition is vacuous. This agrees with the conclusions above: for $K=\mathbb{R}$ the manifold has real dimension $0$, and for $K=\mathbb{C}$ it has complex dimension $0$. [/step]