# Conceptual algebraic proof that Grassmannian is closed in Plucker embedding

I'm planning lectures for my intro algebraic geometry course, and I noted something awkward that is coming up. We're starting projective varieties soon. Of course, we'll prove that projective maps are closed.

Then, I want to talk about Grassmannians. I want to show that the Grassmannian is closed in the Plucker embedding. In other words, I want to know that the set of rank $1$ tensors is closed in $\mathbb{P} \left( \bigwedge\nolimits^k \mathbb{C}^n \right)$. This sounds like it should be a great application of the theorem that projective maps are closed, right? Send $\mathbb{P}^{n-1} \times \mathbb{P}^{n-1} \times \cdots \times \mathbb{P}^{n-1}$ to $\mathbb{P} \left( \bigwedge\nolimits^k \mathbb{C}^n \right)$ by $(v_1, v_2,\ldots, v_k) \mapsto v_1 \wedge v_2 \wedge \cdots \wedge v_k$? Except that this is only a rationally defined map -- it isn't defined when the $v_i$ are linearly dependent.

Of course, I can prove that the rank $1$ tensors are closed by brute force.

I also know some conceptual explanations that are not appropriate as the primary proof for this class (though I may well comment on some of them):

• Working topologically over $\mathbb{C}$, every $k$-plane has an orthonormal basis, so $U(n)$ acts transtively on $G(k,n)$ and $G(k,n)$ is compact.

• In terms of the valuative criterion for closedness, it is enough to take a $k \times n$ matrix with entries valued in a dvr and rank $k$ at the generic point and write down its limit in $G(k,n)$. We can do this explicitly in terms of Smith normal form.

• When a reductive group $G$ acts on an affine variety $X$, the map from the semistable points to Proj of the invariant ring is surjective. Apply this with $G = GL_k$ and $X$ the $k \times n$ matrices. I suspect I am also implicitly using that the ring of invariants is generated by the $k \times k$ minors, which is already nontrivial.

Is there some clever algebraic proof I'm missing, ideally one which uses that projective morphisms are closed?

For context, this is a mixed grad-udergrad course, taught out of Shavarevich volume 1. Everything is over an algebraically closed field and done in a fairly concrete way.

• Doesn't this follow from the Plücker relations? – darij grinberg Sep 18 '14 at 1:58
• @darijgrinberg How do you know that any tensor which obeys the Plucker relations can actually be factored as $v_1 \wedge v_2 \wedge \cdots \wedge v_k$? Before you give too quick an answer, recall that the property of being rank $\leq r$ need not be closed in $U \otimes V \otimes W$, so you have to either use a special property of $\bigwedge$ or a special property of rank $1$. – David E Speyer Sep 18 '14 at 2:09
• @DavidSpeyer: This is fairly well-known. But searching for a nice proof on the internet (the one in my writeup of Etingof's notes is not nice...), I found this, which answers your question just as well: math.rice.edu/~evanmb/math465spring11/math465Grassmannians.pdf (first proof on p. 3). – darij grinberg Sep 18 '14 at 2:26
• Plucker relations are a statement that if you take arbitrary contraction of $w$ by $k-1$ elements of the dual space and then wedge it with $w$, you will get $0$. If you have a tensor $w$ that satisfies this, then consider the space of all such contractions. Take a basis of this space $A$ and complete it to the basis of $V$. Write $w$ as a linear combination of wedges of subsets of this basis. If there is a nonzero coefficient at something other than ones in $A$, then one can get a contraction that is not in $A$. Thus $w$ is up to a scalar the top wedge of $A$. Is this what you need? – Lev Borisov Sep 18 '14 at 3:18
• You'll find a clean proof in Bourbaki's Algebra III, §11, no. 12. – abx Sep 18 '14 at 5:33

Here is the argument I have written up in my thesis. It was suggested to me by my advisor Jarod Alper. We use the fact that a proper monomorphism is a closed immersion (EGA IV, 18.12.6). Furthermore, we will also use the functor of points perspective of the Grassmannian. Now because the Grassmannian is proper and projective space is separated, the Plucker embedding is proper (this is the property $\mathscr{P}$ exercise in Ravi Vakil's notes). Thus we just need to show it's a monomorphism. However, because :

1. The Grassmannian and projective space are covered respectively by the open subfunctors $\textbf{Gr}(d,n)_I$ and $\Bbb{P}(\bigwedge\nolimits^{\!d} \mathcal{O}_{\operatorname{Spec} \Bbb{Z}}^{\oplus n})_I$.
2. An endomorphism of a vector bundle $\varphi : \mathcal{E} \to \mathcal{E}$ is an isomorphism iff $\det \varphi$ is;

it is enough to show the base change $$\textbf{Gr}(d,n)_I \to \Bbb{P}\big(\bigwedge\nolimits^{\!d} \mathcal{O}_{\operatorname{Spec} \Bbb{Z}}^{\oplus n}\big)_I$$ is a monomorphism. One can view fact (2) above as the $\mathcal{O}_X$-module analog of the fact that a matrix is invertible iff its determinant is non-zero.

Now given a scheme $X$ and quotient $\mathcal{O}_{X}^{\oplus n} \twoheadrightarrow \mathcal{F}$, we may always replace it with a quotient $\mathcal{O}_X^{\oplus n} \twoheadrightarrow \mathcal{O}_X^I$ such that the composition $\mathcal{O}_X^I \hookrightarrow \mathcal{O}_X^{\oplus n} \twoheadrightarrow \mathcal{O}_X^I$ is the identity. This follows from the definition of when two quotients are equal in $\textbf{Gr}(d,n)_I(X)$. Henceforth, we will only work with quotients $M : \mathcal{O}_X^{\oplus n} \twoheadrightarrow \mathcal{O}_X^I$ of the following form: $M$ is a matrix (with coefficients in $\Gamma(X,\mathcal{O}_X)$) such that the $d$ columns of $M$ corresponding to the subset $I$ are equal to the columns of the $d \times d$ identity matrix. Now let $X$ be a scheme and suppose there are two quotients $M,M' \in \textbf{Gr}(d,n)_I(X)$ such that $\bigwedge^d M$ and $\bigwedge^d M'$ are equal. This by definition means we have a diagram where the top, bottom rows are the identity and so is the left most column. Thus the displayed isomorphism is an honest equality. But now this means that the $d \times d$ minors of $M$ and $M'$ are equal and hence $M = M'$. In other words, the map $\textbf{Gr}(d,n)_I \to \Bbb{P}(\bigwedge^d \mathcal{O}_{\operatorname{Spec} \Bbb{Z}}^{\oplus n})_I$ is a monomorphism as claimed.

Here are some details for what I think is a variant of Lev Borisov's proposal involving the Plücker relations in the comments. I think if your goal is just to show that the Grassmannian is closed then this is strictly more work than the argument darij links to involving the rank of $\varphi(\omega)$. But in any case the argument avoids the difficulties you mentioned in the comments.

Let $V$ be a finite-dimensional vector space of dimension $n$ and let $\omega \in \bigwedge^k V$. Wedging with $\omega$ gives a map

$$\varphi(\omega) : V \ni v \mapsto \omega \wedge v \in \bigwedge^{k+1} V.$$

Now consider the nondegenerate pairing $\wedge : \bigwedge^k V \otimes \bigwedge^{n-k} V \to \bigwedge^n V$ and pick a nonzero element of $\bigwedge^n V$, or equivalently an identification $\bigwedge^n V \cong 1$ ($1$ the underlying field). This gives us an identification $\bigwedge^k V \cong (\bigwedge^{n-k} V)^{\ast}$. Let $\omega^{\ast} \in (\bigwedge^{n-k} V)^{\ast} \cong \bigwedge^{n-k} V^{\ast}$ be the image of $\omega$ under this identification. Wedging with $\omega^{\ast}$ gives a map

$$\varphi(\omega^{\ast}) : V^{\ast} \ni f \mapsto \omega^{\ast} \wedge f \in \bigwedge^{n-k+1} V^{\ast}.$$

Dualizing this map gives a map

$$\varphi(\omega^{\ast})^{\ast} : \bigwedge^{n-k+1} V \to V$$

which we can now compose with $\varphi(\omega)$, getting a map

$$\omega \wedge \varphi(\omega^{\ast})^{\ast} : \bigwedge^{n-k+1} V \to \bigwedge^{k+1} V.$$

Claim: $\omega$ is a pure tensor iff this map is $0$.

(Expanding this out in a basis gives the Plücker relations, at least if I'm reading these notes correctly.)

Proof. $\Rightarrow$: let $v_1, v_2, \dots v_n$ be a basis with respect to which $\omega = v_1 \wedge \dots \wedge v_k$. Identify $\bigwedge^n V$ with $1$ using $v_1 \wedge \dots \wedge v_n$. Let $v_1^{\ast}, v_2^{\ast}, \dots v_n^{\ast}$ be the dual basis. Then $\omega^{\ast} = v_{k+1}^{\ast} \wedge \dots \wedge v_n^{\ast}$, so the image of $\varphi(\omega^{\ast})^{\ast}$ lies in $\text{span}(v_1, v_2, \dots v_k)$, which wedging with $\omega$ annihilates.

$\Leftarrow$: if $\omega \neq 0$, then $\omega^{\ast} \neq 0$, hence $\varphi(\omega^{\ast}) \neq 0$, hence $\varphi(\omega^{\ast})^{\ast} \neq 0$. Let $v_1, v_2, \dots v_i$ be a basis of $\text{im}(\varphi(\omega^{\ast})^{\ast})$ and complete it to a basis $v_1, v_2, \dots v_n$ of $V$. Again, let $v_1^{\ast}, v_2^{\ast}, \dots v_n^{\ast}$ denote the dual basis.

By hypothesis, $\omega \wedge v_1, \omega \wedge v_2, \dots \omega \wedge v_i = 0$, hence there is some $\omega' \in \bigwedge^{k-i} V$ such that

$$\omega = \omega' \wedge v_1 \wedge \dots \wedge v_i.$$

If $i \ge k$ (in which case this argument shows that $i = k$) then we are done. Otherwise, there must be some $j \in \{ i+1, i+2, \dots n \}$ such that $\omega^{\ast}$ is not divisible by $v_j^{\ast}$, hence such that $\varphi(\omega^{\ast}) v_j^{\ast} \neq 0$. But then there must exist some $\alpha \in \bigwedge^{n-k+1} V$ such that $\alpha(\varphi(\omega^{\ast}) v_j^{\ast}) \neq 0$, hence there must exist some element of $\text{im}(\varphi(\omega^{\ast})^{\ast})$ (namely $\varphi(\omega^{\ast})^{\ast} \alpha$) on which $v_j^{\ast}$ evaluates to something nonzero. But this contradicts $\text{im}(\varphi(\omega^{\ast})^{\ast}) = \text{span}(v_1, v_2, \dots v_i)$.

Hence $i = k$ as desired. $\Box$

• In order for this to be a valid solution one really has to rewrite it to be over rings, or at least local rings, and not just fields, so the language of linear algebra over fields (in which nonzero coefficients are units) is insufficient, strictly speaking. – user27920 Sep 18 '14 at 13:13
• @user52824: that isn't clear to me. The OP specifically says the course works over an algebraically closed field. – Qiaochu Yuan Sep 18 '14 at 17:51
• @QiaochuYuan: You've not understood what I am saying. Even over an algebraically closed field, your solution is incomplete, since all you show is that the image is closed, which is much weaker than showing that the morphism is a closed immersion. You can't prove something is a closed immersion by computing on field-valued points. Your method can be upgraded to work over the dual numbers, etc., to be a correct solution, but as written it is incomplete. (One can also proceed in other ways without using non-reduced rings, by concretely computing on suitable affine charts.) – user27920 Sep 19 '14 at 1:23
• Ah, now I see that there are two ways to read the question: saying the Grassmannian is "closed" could mean just closed image, or the stronger statement about closed immersion. But seriously, in practice it is incredibly important that the latter is true (akin to being a submanifold, not just closed subset), so to not prove this finer result is to shortchange the students; so I think one should use the more refined interpretation of what the question is asking. – user27920 Sep 19 '14 at 1:25
• I think there's a strong argument to be made that in an intro course (esp. with undergrads whose alg. background may be limited), that it's a good idea to only prove that the image is closed. One could supplement that with an explanation of why algebraic geometers are interested in proving the stronger statement that the map is a closed immersion (using the manifold analogy to motivate why) and then mention that in this case one has to work over rings, not just alg. closed fields, and the theory to do this is the theory of schemes, something you should learn in your next alg. geom. course. – Michael Joyce Sep 19 '14 at 15:50

One could argue as follows: if $G$ is a connected algebraic group acting (algebraically) on a vector space $W$, then orbits of $G$ of minimal dimension in ${\mathbb P}(W)$ are closed (by the dominant mapping theorem). Now take $G=GL(V)$ and consider an orbit $O$ of minimal dimension in ${\mathbb P}(\wedge ^ k V)$. The diagonals $T$ in $GL(V)$ have fixed point $p$ in $O$ (an easy version of the Borel fixed point theorem). That is $p$ is an eigenvector for $T$ in $\wedge ^k V$; but the only eigenvectors of $T$ are of the form $p_0= v_1\wedge \cdots \wedge v_k$ where $v_i$ are in the stadrard basis. Hence the orbit through $p_0$ is closed. This orbit is precisely the Grassmannian.

• Here is how I'd rewrite this. Step 1: If $G$ acts on a projective variety $X$, there is a closed orbit. Step 2: Consider $GL(V)$ acting on $\mathbb{P} \bigwedge^k V$; let $X$ be the closed orbit. Step 3: Consider $T$ acting on $X$. Let $O$ be the closed orbit. Step 4: $O$ is closed in $\mathbb{P}\bigwedge^k V$. But explicit computation shows that the only closed $T$ orbits are the obvious fixed points, and the $GL_k$ orbit through one of them is the Grassmannian. I like it. I need to think about whether Step 1 is as straightforward as it seems, but all the rest is elementary. – David E Speyer Sep 18 '14 at 2:48
• @David: if $Gv$ is an orbit, then the complement of $Gv$ in its closure has smaller dimension by the dominant mapping theorem; of course, in a beginning course on algebraic geometry, I do not know if this is proved before one gets to Grassmannians – Venkataramana Sep 18 '14 at 2:58
• I'm not familiar with the term "dominant mapping theorem", but I suspect you mean $\phi: X \to Y$ dominant implies $\phi(X)$ contains a dense open, usually proved as a lemma on the way to Chevalley's theorem. I can rearrange things so that gets done first. At first I was worried about things like $\mathbb{A}^2$ decomposed into the image of $(x,y) \mapsto (x, xy)$ and its complement, where the lowest dimensional constructible set is not closed, but the observation that $G$-invariant sets contain orbits means that $\bar{X} \setminus X$ contains an orbit. – David E Speyer Sep 18 '14 at 10:30
• This only proves closedness of the image of the morphism, not that it is a closed immersion (for which one has to go beyond calculating with field-valued points in this way). – user27920 Sep 18 '14 at 13:15
• @Venkataramana: I had read "is closed" as meaning "map is a closed immersion". I didn't notice (until you pointed it out) that there was a weaker way to interpret the question, since in practice one really wants this stronger property. – user27920 Sep 19 '14 at 1:26

Here is a fairly explicit proof using that projective maps are closed, you may enjoy it if you are following Shavarevich book. It uses the following techniques:

• Projective maps are closed
• Each curve has a normalisation, which is also treated by Shavarevich
• Algebro-geometric computation of limits, that is, working in the local ring of a point
• Algorithm to compute the Hermite normal form of a matrix with coefficients in a euclidean ring

The last algorithm is an adaptation of Gauss elimination algorithm for euclidean rings, it is very cleanly described in Jacobson's Algebra I, see finitely generated modules over a pid.

Let $X = \mathbb{P}^{n-1} \times \mathbb{P}^{n-1} \times \cdots \times \mathbb{P}^{n-1}$ and $X^o$ be the open subset of $X$ consisting of $k$-uples of independant lines and let $Y = \mathbb{P} \left( \bigwedge\nolimits^k \mathbb{C}^n \right)$. I denote by $\phi : X^o \to Y$ the naive map from $X^o$ to $Y$ you described in your question.

## First we construct the rational image of the map

The image $Z$ of $\phi$ is its image as a rational map, we therefore follow the standard construction: let $\Gamma^o\subset X\times Y$ be the graph of $\phi$ and $\Gamma$ its closure in $X\times Y$. We let $Z$ be the projection of $\Gamma$ in $Z$, which is closed because of the theorem you would like to illustrate.

## Then we show that the rational image coincides with the regular image

Let $Z^o = \phi(X^o)$ the image of the map and let us how that $Z^o = Z$ i.e that the Grassmann variety is closed. For any point in $Z$ there is curve $C^o \subset \Gamma^o$ whose closure $C$ meets $X\times\{[z]\}$ — that, is, such that the projection of $C$ contains $[z]$. Let $([v_1], \ldots, [v_k], [z])$ be such a point of $C$.

We now do an infinitesimal analysis around this point to show that $[z]$ belongs to $Z^o$. The key tool of our analysis is the Gauss-Hermite algorithm, which we apply to some matrix with coefficients in local ring of a thickening of $[z]$ in $C$ — or a normalisation of $C$ since we need this ring to be euclidean.

We can find a hyperplane in $\mathbb{P}^{n-1}$ avoiding the various $[v_i]$s, this yields a cell $E = \mathbb{C}^{n-1}$, so that thickening the point $([v_1], \ldots, [v_k], [z])$ yields $k$ vectors $v_i(\epsilon)$ in $E\otimes \mathbb{C}[\epsilon]$ where $\epsilon$ is the local parameter of our thickening.

Applying Gauss-Hermite algorithm yields vectors $\epsilon^{a_i}(w_i +o(\epsilon))$ such that $$v_1(\epsilon) \wedge \cdots \wedge v_j(\epsilon) = \epsilon^{a_1}(w_1 +o(\epsilon))\wedge\cdots\wedge \epsilon^{a_j}(w_j +o(\epsilon))$$ for all $j$. We conclude that $[z] = [w_1\wedge\cdots\wedge w_k]$, that is, $Z^o = Z$.

Instead of taking $(P^{n-1})^k \dashrightarrow P(\Lambda^kC^n)$ you can argue inductively (on $k$) by consider the map $$F(k-1,k) = P_{Gr(k-1,n)}(O^n/U_{k-1}) \to Gr(k,n),$$ where $U_{k-1}$ is the tautological bundle on $Gr(k-1,n)$.

Here is one more approach, I think it is different from the others.

Inside $\mathbb{P} \left( \bigwedge\nolimits^k V \right) \times \mathbb{P} V$, let $X$ denote $\{ (\omega, v) : \omega \wedge v = 0 \}$. This is clearly closed. Consider the projection $X \to \mathbb{P} \left( \bigwedge\nolimits^k V \right)$. Fiber dimension is upper semicontinuous, so let $d-1$ be the greatest fiber dimension and let $\Gamma$ be the closed locus in $\mathbb{P} \left( \bigwedge\nolimits^k V \right)$ where that greatest dimension is achieved.

Looking at a rank one point of $\mathbb{P} \left( \bigwedge\nolimits^k V \right)$, $d \geq k$. For $x \in \Gamma$, let $V_x \subset V$ be the subspace for which $\mathbb{P}(V)$ is the fiber. So $d = \dim V_x$, choose a basis $v_1$, $v_2$, ..., $v_d$. Then $v_1 \wedge \cdots \wedge v_d$ divides $x$, as in several other proofs, so $d \leq k$. This shows that $d=k$ and that every point of $\Gamma$ is of rank $1$, conversely, it is obvious that the fiber above every rank one point is of dimension $k-1$.

Basically, this is Qiaochu's answer/Lev Borsiov's comment, except using semicontinuity of fiber dimension instead of explicit equations.

• This is not a solution because it proves closedness of the image of the morphism but not that the morphism is a closed immersion. If it were reworked to take place over rings then one might be able to extract a correct solution along these lines, but Ben Lim's argument is the "right" one over rings (from which the Plucker relations also drop out in the correct "scheme-theoretic" generality; it is all in Grothendieck's Seminaire Cartan exposes). – user27920 Sep 18 '14 at 13:17
• @user52824 Note my answer above works over any base scheme $S$; I don't assume anything about working over a field or anything like that. However, I believe it is enough to work over the integers because Grassmannians "base change correctly". – Ben Lim Sep 18 '14 at 23:57
• @BenLim: I completely agree -- your solution is the right one, and is valid over any base scheme (and adapts to work in other geometric categories: complex-analytic spaces, rigid-analytic spaces, etc). – user27920 Sep 19 '14 at 1:27

In the paper

• Mike Eastwood, Peter W. Michor: Some remarks on the Plücker relations. Rendiconti del Circolo Matematico di Palermo, Serie II, Suppl. 63 (2000), 85-88. pdf of a longer version

there are 5 versions of the Pluecker relations in theorem 1. Versions (3) and (4) might be helpful. All proofs are included.