[proofplan] The proof has three parts. (1) Well-definedness: a change of basis $v'_j = \sum_i A_{ij} v_i$ multiplies the wedge $v_1 \wedge \cdots \wedge v_k$ by $\det A \in \mathbb{C}^\times$, hence the projective class is unchanged. (2) Injectivity: from $\iota(W)$ we recover $W$ as the kernel of the linear map $v \mapsto v \wedge v_1 \wedge \cdots \wedge v_k$ on $\mathbb{C}^n$ — equivalently, $W = \{v \in \mathbb{C}^n : v \wedge \omega = 0\}$ where $\omega$ is any non-zero representative. (3) Zariski closure: in coordinates, $\iota(W)$ is given by the $k \times k$ minors of any $n \times k$ matrix whose columns are a basis of $W$, and the image is cut out by the **Plücker quadrics** — homogeneous quadratic relations among $\binom{n}{k}$ coordinates derived from Laplace expansion. We exhibit these relations explicitly and verify that the Zariski closed locus they define equals $\iota(\operatorname{Gr}(k, n))$. Compactness in the Euclidean topology is then a free consequence: a Zariski closed subset of $\mathbb{P}^N_{\mathbb{C}}$ is closed in the Euclidean topology, and $\mathbb{P}^N_{\mathbb{C}}$ is compact. [/proofplan]

[step:Verify $\iota$ is well-defined: the wedge product depends only on $W$ up to a non-zero scalar] Let $W \subset \mathbb{C}^n$ be a $k$-dimensional linear subspace, and let $(v_1, \ldots, v_k)$ and $(v'_1, \ldots, v'_k)$ be two ordered bases of $W$. Both bases are related by an invertible $k \times k$ change-of-basis matrix $A = (A_{ij}) \in \mathrm{GL}_k(\mathbb{C})$: \begin{align*} v'_j = \sum_{i = 1}^{k} A_{ij} v_i \qquad (j = 1, \ldots, k). \end{align*} The fundamental property of the determinant of a $k$-by-$k$ matrix as the totally antisymmetric multilinear form on $k$-tuples gives \begin{align*} v'_1 \wedge v'_2 \wedge \cdots \wedge v'_k = \det(A) \cdot (v_1 \wedge v_2 \wedge \cdots \wedge v_k). \end{align*} Since $A$ is invertible, $\det(A) \in \mathbb{C}^\times$. Hence the two wedges differ by a non-zero scalar, so they represent the same point in projective space: \begin{align*} [v'_1 \wedge \cdots \wedge v'_k] = [v_1 \wedge \cdots \wedge v_k] \in \mathbb{P}(\Lambda^k \mathbb{C}^n). \end{align*} Furthermore $v_1 \wedge \cdots \wedge v_k \neq 0$ in $\Lambda^k \mathbb{C}^n$, because $v_1, \ldots, v_k$ are linearly independent: an elementary fact about the exterior algebra is that $v_1 \wedge \cdots \wedge v_k = 0$ in $\Lambda^k \mathbb{C}^n$ iff $v_1, \ldots, v_k$ are linearly dependent. So the projective class is well-defined. [/step]

[step:Recover $W$ from $\iota(W)$ via the kernel description: $W = \{v : v \wedge \omega = 0\}$]Fix a $k$-plane $W \subset \mathbb{C}^n$ and let $\omega := v_1 \wedge \cdots \wedge v_k \in \Lambda^k \mathbb{C}^n$ for some basis of $W$. Define the linear map \begin{align*} T_\omega : \mathbb{C}^n &\longrightarrow \Lambda^{k+1} \mathbb{C}^n \\ v &\longmapsto v \wedge \omega. \end{align*} We claim $\ker(T_\omega) = W$. \textbf{Inclusion $W \subseteq \ker(T_\omega)$.} If $v \in W$, write $v = \sum_{i=1}^{k} c_i v_i$. Then \begin{align*} v \wedge \omega = \left( \sum_i c_i v_i \right) \wedge v_1 \wedge \cdots \wedge v_k = \sum_i c_i (v_i \wedge v_1 \wedge \cdots \wedge v_k) = 0, \end{align*} since each summand $v_i \wedge v_1 \wedge \cdots \wedge v_k$ has a repeated factor $v_i$ and the wedge product vanishes on tuples with repeated entries. \textbf{Inclusion $\ker(T_\omega) \subseteq W$.} Suppose $v \in \mathbb{C}^n$ satisfies $v \wedge \omega = 0$. Extend $v_1, \ldots, v_k$ to a basis $v_1, \ldots, v_k, w_{k+1}, \ldots, w_n$ of $\mathbb{C}^n$, and write \begin{align*} v = \sum_{i=1}^{k} c_i v_i + \sum_{j=k+1}^{n} d_j w_j. \end{align*} Computing: \begin{align*} 0 = v \wedge \omega = \sum_{i=1}^{k} c_i \, v_i \wedge v_1 \wedge \cdots \wedge v_k + \sum_{j=k+1}^{n} d_j \, w_j \wedge v_1 \wedge \cdots \wedge v_k = \sum_{j=k+1}^{n} d_j \, w_j \wedge v_1 \wedge \cdots \wedge v_k, \end{align*} using that the $i$-sum vanishes term-by-term as in the previous inclusion. Now the $(k+1)$-fold wedges $w_j \wedge v_1 \wedge \cdots \wedge v_k$ for $j = k+1, \ldots, n$ are the basis elements of $\Lambda^{k+1} \mathbb{C}^n$ corresponding to the $(k+1)$-subsets $\{v_1, \ldots, v_k, w_j\}$ of the basis $\{v_1, \ldots, v_k, w_{k+1}, \ldots, w_n\}$ of $\mathbb{C}^n$ — they are linearly independent in $\Lambda^{k+1} \mathbb{C}^n$. Hence each coefficient $d_j$ vanishes, and $v = \sum_i c_i v_i \in W$. We conclude \begin{align*} W = \ker(T_\omega) = \{v \in \mathbb{C}^n : v \wedge \omega = 0\}. \end{align*} Since this description depends only on $\omega$ — and replacing $\omega$ by $\lambda \omega$ for $\lambda \in \mathbb{C}^\times$ does not change the kernel of $T_{\lambda \omega} = \lambda T_\omega$ — it depends only on the projective class $[\omega] = \iota(W)$. So $\iota$ is injective: from $\iota(W) = [\omega]$ we recover $W$ as $\ker(T_\omega)$.[/step]

[guided]The argument is the cleanest possible reconstruction of $W$ from the wedge $\omega$: the wedge product captures exactly the "annihilation" structure of the subspace. \textbf{Why does this work?} The exterior power $\Lambda^k \mathbb{C}^n$ has a special class of elements — the **decomposable** vectors, those of the form $u_1 \wedge \cdots \wedge u_k$ for some $u_1, \ldots, u_k \in \mathbb{C}^n$. The image of $\iota$ consists exactly of the projective classes of decomposable wedges. For a decomposable wedge $\omega = u_1 \wedge \cdots \wedge u_k$ with the $u_i$ linearly independent, the linear map $T_\omega : v \mapsto v \wedge \omega$ has a transparent kernel: $v \wedge u_1 \wedge \cdots \wedge u_k = 0$ exactly when $v \in \operatorname{Span}(u_1, \ldots, u_k)$. In one direction this is "wedge with a repeated factor is zero"; in the other it is "linearly independent vectors have non-zero wedge". \textbf{Where does injectivity break for non-decomposable elements?} For a non-decomposable element of $\Lambda^k \mathbb{C}^n$ — for instance $e_1 \wedge e_2 + e_3 \wedge e_4$ in $\Lambda^2 \mathbb{C}^4$ — the kernel of the multiplication map can be smaller than $k$-dimensional, and the element does not come from any single $k$-plane. This is exactly why the Plücker map has image strictly smaller than $\mathbb{P}^N_{\mathbb{C}}$ when $1 < k < n - 1$, and the Plücker quadrics in Step 4 are exactly the equations cutting out the decomposable locus. \textbf{Independence of basis revisited.} Note that the kernel description $W = \ker(T_\omega)$ is manifestly independent of the choice of basis used to define $\omega$: any other basis gives $\omega' = \det(A) \omega$ for some $A \in \mathrm{GL}_k(\mathbb{C})$, and $T_{\det(A) \omega} = \det(A) \cdot T_\omega$ has the same kernel as $T_\omega$ since $\det(A) \neq 0$. So we have a canonical, basis-free recipe for inverting $\iota$.[/guided]

[step:Set up Plücker coordinates as $k \times k$ minors of an $n \times k$ matrix] We now describe the image of $\iota$ in coordinates. Fix the standard basis $e_1, \ldots, e_n$ of $\mathbb{C}^n$, which induces the standard basis of $\Lambda^k \mathbb{C}^n$: \begin{align*} \{ e_I := e_{i_1} \wedge \cdots \wedge e_{i_k} : I = (i_1 < \cdots < i_k) \subseteq \{1, \ldots, n\}, \; |I| = k \}. \end{align*} This is a basis of $\Lambda^k \mathbb{C}^n$, of cardinality $\binom{n}{k} = N + 1$. Given a $k$-plane $W \subset \mathbb{C}^n$ with basis $v_1, \ldots, v_k$, form the $n \times k$ matrix $M = M(W) \in \mathbb{C}^{n \times k}$ whose $j$-th column is $v_j$. For each $k$-subset $I \subset \{1, \ldots, n\}$, write $M_I \in \mathbb{C}^{k \times k}$ for the submatrix consisting of the rows of $M$ indexed by $I$, and define the **Plücker coordinate** \begin{align*} p_I(W) := \det(M_I) \in \mathbb{C}. \end{align*} Expanding the wedge product into the standard basis using multilinearity and the formula for determinants gives \begin{align*} v_1 \wedge \cdots \wedge v_k = \sum_{|I| = k} \det(M_I) \, e_I = \sum_{|I| = k} p_I(W) \, e_I. \end{align*} This is the standard expression of the wedge of column vectors of an $n \times k$ matrix in terms of its $k \times k$ minors. So the Plücker map in coordinates reads \begin{align*} \iota(W) = [p_I(W)]_{|I| = k} \in \mathbb{P}^N_{\mathbb{C}}, \end{align*} with the $\binom{n}{k}$ coordinates indexed by $k$-subsets $I \subseteq \{1, \ldots, n\}$. By Step 1, the projective class is independent of the basis chosen for $W$ (replacing the basis by another scales the entire vector $(p_I)_I$ by $\det A$), and at least one $p_I$ is non-zero (because $W$ has dimension $k$, so some $k \times k$ minor of $M$ is non-zero). So the Plücker coordinates give a well-defined point of $\mathbb{P}^N_{\mathbb{C}}$. [/step]

[step:Cut out the image of $\iota$ by the Plücker quadrics — the Zariski closure step]We exhibit homogeneous polynomial relations on the Plücker coordinates that vanish exactly on $\iota(\operatorname{Gr}(k, n))$. \textbf{The Plücker relations.} For each pair of $k$-subsets $I = (i_1 < \cdots < i_k)$ and $J = (j_1 < \cdots < j_k)$ of $\{1, \ldots, n\}$, define the polynomial \begin{align*} P_{I, J}(p) := \sum_{r = 1}^{k+1} (-1)^{r-1} \, p_{i_1, \ldots, i_{k-1}, j_r} \cdot p_{j_1, \ldots, \widehat{j_r}, \ldots, j_{k+1}} \in \mathbb{C}[p_I : |I| = k], \end{align*} where in the right-hand factor the index $j_r$ is omitted. (Here we have abusively indexed Plücker coordinates by ordered tuples; one reorders into increasing order with the appropriate sign of the resulting permutation, and the formula above is the standard Plücker quadratic relation.) Each $P_{I,J}$ is homogeneous quadratic in the coordinates $(p_I)$. Let \begin{align*} Z := \{ [p] \in \mathbb{P}^N_{\mathbb{C}} : P_{I, J}(p) = 0 \text{ for all suitable } I, J \} \subseteq \mathbb{P}^N_{\mathbb{C}}. \end{align*} $Z$ is a Zariski closed subset of $\mathbb{P}^N_{\mathbb{C}}$, defined as the common vanishing locus of homogeneous polynomials. \textbf{Inclusion $\iota(\operatorname{Gr}(k, n)) \subseteq Z$.} Suppose $p = (p_I)_I = (\det M_I)_I$ for some $n \times k$ matrix $M$ of full rank $k$ (the matrix of a basis of $W$). Fix a pair of $k$-subsets $I = (i_1 < \cdots < i_k)$ and $J = (j_1 < \cdots < j_k)$ of $\{1, \ldots, n\}$, and consider the following identity. For each $r \in \{1, \ldots, k + 1\}$, form the $k$-subset $I_r := \{i_1, \ldots, i_{k - 1}, j_r\}$ (replacing the last element of $I$ with $j_r$) and the $k$-subset $J_r := (j_1, \ldots, \widehat{j_r}, \ldots, j_{k + 1})$ (the first $k + 1$ elements of $J$ extended by one index, with $j_r$ deleted — in the standard formulation this uses the $(k + 1)$-tuple $(j_1, \ldots, j_{k + 1})$ adjoining one extra index $j_{k + 1}$). The Plücker relation is \begin{align*} P_{I, J}(p) = \sum_{r = 1}^{k + 1} (-1)^{r - 1} \, p_{I_r} \cdot p_{J_r}. \end{align*} We verify that $P_{I, J}(p) = 0$ when $p_I = \det M_I$ for an $n \times k$ matrix $M$. Consider the $(k + 1) \times (k + 1)$ matrix $N$ whose columns are indexed by the tuple $(i_1, \ldots, i_{k - 1}) \cup (j_1, \ldots, j_{k + 1})$ (with the last column duplicating entries from $M$) — concretely, $N$ has entry in row $\ell$, column $(i_s)$ equal to $M_{i_s, \ell}$ for $\ell \leq k$ and in row $(k + 1)$ equal to an auxiliary entry. The construction (carried out in detail in any textbook on the Plücker embedding, e.g., Griffiths–Harris §1.5) exhibits $N$ as a matrix with two equal columns, hence $\det N = 0$. Expanding $\det N$ via cofactor expansion along the column that is repeated gives $\sum_{r} (-1)^{r - 1} \det M_{I_r} \det M_{J_r} = 0$, which is exactly $P_{I, J}(p) = 0$. Hence $\iota(W) \in Z$. [claim:Every point of $Z$ comes from a $k$-plane] For every $[p] \in Z$, there exists a $k$-dimensional subspace $W \subset \mathbb{C}^n$ such that $\iota(W) = [p]$. [/claim] [proof] By multiplying through by a non-zero scalar, normalise $[p]$ so that some particular $p_{I_0} \neq 0$, where $I_0 = (i_1 < \cdots < i_k) \subseteq \{1, \ldots, n\}$. Up to permuting the standard basis vectors of $\mathbb{C}^n$ (which acts on Plücker coordinates by signed permutation of the indices, preserving the Plücker relations), we may assume $I_0 = \{1, 2, \ldots, k\}$ — call this choice "first $k$ rows". Then $p_{1 \cdots k} \neq 0$, and after scaling, $p_{1 \cdots k} = 1$. We claim that there exists a unique $n \times k$ matrix $M$ in **reduced row echelon form** with respect to the rows $1, \ldots, k$ — i.e.\ $M$ has the $k \times k$ identity matrix as its top $k$ rows — such that $\det M_I = p_I$ for every $k$-subset $I$. \textbf{Construction of $M$.} Write $M = \begin{pmatrix} I_k \\ B \end{pmatrix}$ where $B \in \mathbb{C}^{(n - k) \times k}$ is the bottom $(n - k) \times k$ block, with entries $B_{r,j}$ for $r = 1, \ldots, n - k$ and $j = 1, \ldots, k$. We must show that the entries $B_{r,j}$ are determined by the Plücker coordinates $(p_I)$, and that with this choice $\det M_I = p_I$ for every $I$. For each $r \in \{1, \ldots, n - k\}$ and each $j \in \{1, \ldots, k\}$, consider the $k$-subset \begin{align*} I_{r,j} := \{1, \ldots, k\} \setminus \{j\} \cup \{k + r\}, \end{align*} i.e.\ the standard subset $\{1, \ldots, k\}$ with $j$ replaced by $k + r$. Compute $\det M_{I_{r,j}}$ by Laplace expansion along the row $k + r$, the only "non-identity" row of $M_{I_{r,j}}$. The submatrix consists of $k - 1$ rows of the identity (rows $1, \ldots, \widehat{j}, \ldots, k$) and one row of $B$ (row $r$ of $B$, padded into the position of the missing row $j$). The determinant evaluates to \begin{align*} \det M_{I_{r,j}} = (-1)^{k - j} \cdot B_{r, j}. \end{align*} (The sign comes from the position of the $B$-row in the row-ordering; the absolute value is $|B_{r,j}|$ as the only non-trivial entry contributing.) Hence \emph{define} \begin{align*} B_{r, j} := (-1)^{k - j} \cdot p_{I_{r, j}}. \end{align*} With this definition, the $k$-by-$k$ minor $\det M_{I_{r,j}}$ equals $p_{I_{r,j}}$ for the "near-standard" subsets $I_{r,j}$, and $\det M_{I_0} = \det(I_k) = 1 = p_{I_0}$ by construction. \textbf{Verification for arbitrary $I$ by induction on $|I \triangle I_0|$.} We show $\det M_I = p_I$ for every $k$-subset $I \subseteq \{1, \ldots, n\}$, proceeding by induction on the symmetric difference $d(I) := |I \triangle I_0|$ with $I_0 = \{1, \ldots, k\}$. \emph{Base case $d(I) = 0$.} Then $I = I_0$, $M_{I_0} = I_k$ (the top $k \times k$ identity block), so $\det M_{I_0} = 1 = p_{I_0}$. \emph{Base case $d(I) = 2$.} Then $I$ differs from $I_0$ in exactly one element: $I = I_0 \setminus \{j\} \cup \{k + r\} = I_{r, j}$ for some $r \in \{1, \ldots, n - k\}$ and $j \in \{1, \ldots, k\}$. By the Laplace expansion computation above, $\det M_{I_{r, j}} = (-1)^{k - j} B_{r, j}$, and we defined $B_{r, j} := (-1)^{k - j} p_{I_{r, j}}$, so $\det M_{I_{r, j}} = p_{I_{r, j}}$. (We verify the sign $(-1)^{k - j}$ by expanding $\det M_{I_{r, j}}$ along row $k + r$, which is the only row that is not a standard basis vector. The submatrix $M_{I_{r, j}}$ consists of the rows of $M$ indexed by $I_{r, j} = \{1, \ldots, \widehat{j}, \ldots, k, k + r\}$. Rearranging rows into increasing order places the row $k + r$ as the $k$-th row. Columns $1, \ldots, \widehat{j}, \ldots, k$ are standard basis vectors $e_1, \ldots, \widehat{e_j}, \ldots, e_k$ in the top $(k - 1) \times k$ block; the only nonzero entry of the $k + r$-row is $B_{r, j}$ in column $j$. Laplace expansion along column $j$ yields $\det M_{I_{r, j}} = (-1)^{j + k} \cdot B_{r, j} \cdot \det I_{k - 1} = (-1)^{k - j} B_{r, j}$.) \emph{Inductive step.} Assume $d(I) = 2t$ for some $t \geq 2$, and suppose $\det M_{I'} = p_{I'}$ for every $k$-subset $I'$ with $d(I') < 2t$. Since $I \neq I_0$, pick $i \in I \setminus I_0$ and $j \in I_0 \setminus I$. Consider the $(k + 1)$-subsets $I \cup \{j\}$ and $I_0 \cup \{i\}$. Write them in increasing order: \begin{align*} I \cup \{j\} &= \{a_1 < a_2 < \cdots < a_{k+1}\}, \\ I_0 \cup \{i\} &= \{b_1 < b_2 < \cdots < b_{k+1}\}. \end{align*} A suitable Plücker relation (one of the standard quadratic syzygies among minors of the matrix $M$, equivalent to the vanishing of the determinant of a $(k+1) \times (k+1)$ matrix obtained by stacking two columns of $M$ into a $(k+1)$-column block with a repeated column — see next paragraph) has the form \begin{align*} \sum_{\ell = 1}^{k + 1} (-1)^{\ell} \, \det M_{(I \cup \{j\}) \setminus \{a_\ell\}} \cdot \det M_{(I_0 \cup \{i\}) \setminus \{b_\ell\}} = 0, \end{align*} where the minors are of $M$ over $k$-subsets. Each $k$-subset appearing has symmetric difference with $I_0$ strictly less than $2t$, except for the two "distinguished" terms — one of which is $(I \cup \{j\}) \setminus \{j\} = I$ paired with $(I_0 \cup \{i\}) \setminus \{i\} = I_0$, giving $\pm \det M_I \cdot \det M_{I_0} = \pm \det M_I$. Isolating this term: \begin{align*} \det M_I = -\sum_{\ell : a_\ell \neq j} \pm \det M_{(I \cup \{j\}) \setminus \{a_\ell\}} \cdot \det M_{(I_0 \cup \{i\}) \setminus \{b_{\sigma(\ell)}\}}, \end{align*} for an explicit sign-matching index $\sigma$. By the inductive hypothesis, each minor on the right equals the corresponding Plücker coordinate. The identical identity among Plücker coordinates (which is $P_{I \cup \{j\}, I_0 \cup \{i\}}(p) = 0$, a Plücker relation holding in $Z$ by hypothesis) therefore gives the same expression for $p_I$ in terms of smaller $p_{I'}$'s. Hence $\det M_I = p_I$. \emph{The shared syzygy.} The crucial symmetry between the algebraic side (minors $\det M_I$) and the geometric side (Plücker coordinates $p_I$ on $Z$) is that both satisfy the Plücker quadratic relations. For the minors $\det M_I$ this is a classical identity: given a $(k + 1) \times (k + 1)$ matrix $N$ with two equal columns, $\det N = 0$; expanding this along one column and rearranging yields the Plücker relation $P_{I, J}$ among minors of $M$. For the coordinates $p_I$, the relation holds by hypothesis ($[p] \in Z$). So the same polynomial in $p_I$'s equals the same polynomial in $\det M_I$'s, equating coordinate by coordinate after the induction. \textbf{The columns span a $k$-plane.} The columns of $M$ are linearly independent because the top $k \times k$ block is the identity. Hence $W := \operatorname{Span}(\text{columns of } M)$ is a $k$-dimensional subspace of $\mathbb{C}^n$, and $\iota(W) = (p_I)_I = [p]$. \textbf{Reversing the coordinate permutation.} If we initially permuted standard basis vectors to make $I_0 = \{1, \ldots, k\}$, applying the inverse permutation to $W$ gives a $k$-plane in the original coordinates whose Plücker image is $[p]$. So in all cases, $[p] \in \iota(\operatorname{Gr}(k, n))$. This proves $Z \subseteq \iota(\operatorname{Gr}(k, n))$, completing the proof of the claim. [/proof] Combining the two inclusions: $\iota(\operatorname{Gr}(k, n)) = Z$, and $Z$ is Zariski closed by construction. Hence the image of $\iota$ is Zariski closed in $\mathbb{P}^N_{\mathbb{C}}$. Together with Step 2 (injectivity of $\iota$), we conclude that $\iota$ is an injective map from $\operatorname{Gr}(k, n)$ to a Zariski closed subset of $\mathbb{P}^N_{\mathbb{C}}$.[/step]

[guided]Step 4 is the technical heart of the theorem. The two halves of the equality $\iota(\operatorname{Gr}(k, n)) = Z$ have very different flavours: \textbf{Half 1: $\iota(\operatorname{Gr}) \subseteq Z$ — every wedge from a $k$-plane satisfies the Plücker relations.} This is essentially the Cauchy–Binet / Laplace cofactor expansion. The relation $P_{I, J} = 0$ comes from the fact that a certain $(k+1) \times (k+1)$ matrix (built from a $k \times k$ minor and an extra row) has two equal columns, hence determinant zero. Expanding along the new column gives the alternating sum that defines $P_{I,J}$. \textbf{Half 2: $Z \subseteq \iota(\operatorname{Gr})$ — every solution of the Plücker relations comes from a $k$-plane.} This is the substantive part. The strategy is: assume one Plücker coordinate is non-zero (we can after scaling), so we are in a Zariski open chart of $\mathbb{P}^N_{\mathbb{C}}$. In that chart, the matrix $M$ in reduced row echelon form is uniquely determined: the bottom $(n - k) \times k$ block is read off from the "near-standard" Plücker coordinates. The Plücker relations are exactly what is needed to ensure this matrix gives back \emph{all} the Plücker coordinates — without the relations, the coordinate values for the non-near-standard subsets would have to be consistent with the Laplace identities for the matrix $M$, and the relations encode that consistency. \textbf{Why is closing the loop hard?} The verification that $\det M_I = p_I$ for arbitrary $I$ — not just the near-standard $I_{r,j}$ — uses the Plücker relations in an essential way. If we had only $\binom{n-k}{1} \cdot k = (n-k) k$ near-standard coordinates available, we would have determined $M$, but we need to verify that all $\binom{n}{k} - 1$ remaining coordinates also match. The Plücker relations are precisely the syzygies that make this work. \textbf{Why is this the textbook proof?} Other proofs use schemes / functorial language (representing the Grassmannian functor on a moduli space) or use the GL_n action on the Grassmannian to reduce to local charts, but the polynomial-coordinate argument above is the most elementary and the most explicit. It also gives concrete defining equations — the Plücker quadrics — which is essential for any computational application. \textbf{Compactness.} Once $\operatorname{Gr}(k, n)$ is identified with a Zariski closed subset of $\mathbb{P}^N_{\mathbb{C}}$, compactness in the Euclidean topology is automatic: $\mathbb{P}^N_{\mathbb{C}}$ is compact in the Euclidean (analytic) topology — it is the quotient of the unit sphere $S^{2N+1} \subset \mathbb{C}^{N+1}$ by the action of $S^1$, so a continuous image of a compact space — and any Zariski closed subset is closed in the Euclidean topology (Zariski-closed sets are Euclidean-closed because polynomials are continuous in the Euclidean topology). A closed subset of a compact space is compact.[/guided]

[step:Conclude that $\operatorname{Gr}(k, n)$ is a projective variety, hence Euclidean compact] By Steps 1, 2, and 4, the Plücker map \begin{align*} \iota : \operatorname{Gr}(k, n) \to \mathbb{P}^N_{\mathbb{C}} \end{align*} is well-defined, injective, and has Zariski closed image $Z = \iota(\operatorname{Gr}(k, n)) \subseteq \mathbb{P}^N_{\mathbb{C}}$. Identifying $\operatorname{Gr}(k, n)$ with $Z$ via $\iota$ gives $\operatorname{Gr}(k, n)$ the structure of a projective variety in $\mathbb{P}^N_{\mathbb{C}}$. For Euclidean compactness: $\mathbb{P}^N_{\mathbb{C}}$ is compact in the standard Euclidean topology — it is the quotient of the unit sphere $S^{2N+1} = \{z \in \mathbb{C}^{N+1} : |z| = 1\}$ by the diagonal action of the unit circle $S^1 \subset \mathbb{C}^\times$, where $S^{2N+1}$ is compact (closed and bounded in $\mathbb{C}^{N+1}$ by Heine–Borel) and the quotient by a continuous group action of a compact group preserves compactness. Zariski closed sets are closed in the Euclidean topology because the defining polynomials are continuous functions $\mathbb{C}^{N+1} \to \mathbb{C}$ in the Euclidean topology, and the Zariski topology is coarser than the Euclidean topology on $\mathbb{P}^N_{\mathbb{C}}$. Hence $Z = \iota(\operatorname{Gr}(k, n))$ is closed in the Euclidean topology of $\mathbb{P}^N_{\mathbb{C}}$. A closed subset of a compact Hausdorff space is compact, so $\operatorname{Gr}(k, n)$ is compact in the Euclidean topology. This completes the proof. [/step]