Let $V\subseteq\mathbb{C}^n$ be a variety given as the zero set of some homogeneous polynomials and let $m\leq n$. I am interested in the dimension of the variety $$ W=\{(v_1,\dots,v_m)\in\mathbb{C}^{n\times m}\mid\mbox{ there exists }0\neq (\lambda_1,\dots,\lambda_m)\in\mathbb{C}^m, \lambda_1 v_1+\dots+\lambda_m v_m\in V\}. $$

Now, there is a map $$ \varphi:(\mathbb{C}^m-\{0\})\times\mathbb{C}^{n\times m}\rightarrow\mathbb{C}^n $$ mapping $(\lambda_1,\dots,\lambda_m,v_1,\dots,v_m)$ to $\sum_{i=1}^m \lambda_i v_i$. Then, $W$ is the projection of $\varphi^{-1}(V)$ to $\mathbb{C}^{n\times m}$.

Since $\varphi$ is linear in the variables $\lambda_i$, for any $0\neq \lambda=(\lambda_1,\dots,\lambda_m)$, the fiber of $\lambda$ in $\varphi^{-1}(V)$ is isomorphic to $V\times\mathbb{C}^{n\times (m-1)}$. Thus, the codimension of $W$ should be codim$(V)-m+1$. However, this is not very rigorous. I am also having problems to show that $W$ is actually Zariski closed. By some analysis arguments, I can show that $W$ is closed with respect to the Euclidean topology and since $W$ is the projection of a Zariski closed set, I think $W$ must be Zariski closed. But again, this is not very rigorous.

Any help is appreciated.

  • $\begingroup$ For the codimension, it should be possible to make this into a local argument: The Krull dimension of a Zariski locally closed set, that is a manifold, equals the complex manifold dimension. For Zariski-closedness you have to show, that $W$ can be written as the solution-set of polynomial equations. A priori an image of a polynomial map is the solution-set of polynomial equations and non-equations, so you are left to eliminate the non-equations. $\endgroup$ – Julian Quast Dec 1 '19 at 20:52
  • $\begingroup$ The fact that your polynomial, and condition are both homogeneous means you can projectivise everything, and then the fact that projective varieties are universally closed may be helpful. $\endgroup$ – Chris H Dec 2 '19 at 6:22

Let's start with whether $W$ is closed or not. You're right to be suspicious about the claim that the projection of a closed set is closed: it's not always true, as you can see from projecting $V(xy-1)\subset \Bbb A^2$ to the $x$-axis.

On the other hand, there is such a definition that lets you conclude things like this. We call a variety $X$ universally closed if for any other variety $Y$ the projection $X\times Y\to Y$ is closed. A wide class of such varieties is given by projective varieties, and this suggests a strategy: if we can somehow tinker with what we've got going on in the question to make the appropriate things projective, then maybe we can get a good solution.

First, we note that $A=\varphi^{-1}(V)$ is invariant under nonzero scaling in both factors as a subvariety of $(\Bbb C^m\setminus \{0\}) \times \Bbb C^{n\times m}$: if we scale the $\lambda_i$ by $r$ and the $v_i$ by $s$, then $\sum \lambda_iv_i$ scales by $rs$, and as the equations cutting out $V$ are homogeneous, they evaluate to zero on this scaled vector iff they evaluate to zero on the unscaled vector. This means we can projectivize both factors of the product $(\Bbb C^m\setminus \{0\}) \times \Bbb C^{n\times m}$ and have $\varphi^{-1}(V)$ define a closed subset $Z$ of $\Bbb P(\Bbb C^m)\times \Bbb P(\Bbb C^{nm})$, which after projecting to $\Bbb P(\Bbb C^{nm})$ is again closed. But then $W$ is the affine cone over $\pi(Z)$, the projection of $Z$, so it's closed.

For clarity of notation, we add a subscript to $W$ to make it clear which copy of $\Bbb C^{nm}$ we're working in: $W_m$ is the subset of $\Bbb C^{nm}$ satisfying the requested conditions. Now on to the dimension calculations.

The dimension calculations for $W$ you've presented are not correct. The dimension of $W$ should be sensitive to $m$ and $\dim V$ as the following example shows: let $V$ be the $xy$ 2-plane in $\Bbb C^4$. Then for $m=1$, $W\subset \Bbb C^{nm}$ is just $V$ again and has codimension 2. For $m=2$, $V$ is the zero locus of the determinant of $$\begin{pmatrix} v_{11} & v_{21} & 1 & 0 \\ v_{12} & v_{22} & 0 & 1 \\ v_{13} & v_{23} & 0 & 0 \\ v_{14} & v_{24} & 0 & 0 \end{pmatrix},$$ or $V(v_{13}v_{24}-v_{23}v_{14})$ which is codimension one. For $m=3$, every collection of $v_1,v_2,v_3$ meet the stated criteria (either they're linearly dependent in which case the linearly dependence relation shows that they're in $W$, or they're linearly indepedent and their span is a 3-plane, which must intersect a 2-plane nontrivially), so $W$ is the whole space (similarly for $m=4$). This suggests the following: $\operatorname{codim} W = \max(\operatorname{codim} V - m+1,0)$. This is in fact the case.

First, we may observe that $\dim W_1=\dim V$ and that for any collection of $m$ linearly independent vectors in $(\Bbb C^n)^m$ spanning an $m$ plane $P_m$, we get that the dimension of the intersection $P_m\cap V$ must be at least one (and thus this set of vectors lies in $W_m$) if the dimensions of $V$ and $P_m$ add to more than $n$, which is equivalent to $0\geq \operatorname{codim} V -m+1$. As the condition that $m$ vectors are linearly independent is an open condition inside the irreducible space $\Bbb C^{nm}$ and $W_m$ is a closed subset thereof, this implies that $W_m=\Bbb C^{nm}$ if the given condition is satisfied.

Now we want to show that $\operatorname{codim} W_m > \operatorname{codim} W_{m+1}$. If we can do this, then we'll have our conclusion: the only way to organize a strictly decreasing sequence of integers with the first term $\operatorname{codim} V$ and the $(1+\operatorname{codim} V)^{th}$ term $0$ is the function we described above.

We'll do this by carefully examining the number of lines in general position inside $W$ through a given point in $W$. Pick $m$ so that $W_m$ is not the whole space $\Bbb C^{nm}$. Take a point in $W_m$ where the vectors $v_1,\cdots,v_m$ are linearly independent (this exists: pick $v_1\in V$ and then select linearly independent vectors). Then $(v_1,\cdots,v_m,0)$ is in $W_{m+1}$ via the same vector of $\lambda$s as $(v_1,\cdots,v_m)$ with a $1$ appended at the end. Clearly we can add any multiples of $v_1,\cdots,v_m$ to the final entry via correcting in the $\lambda_i$ terms to make the result $\sum \lambda_iv_i$ still lie in $V$, so $\dim W_{m+1}$ is at least $\dim W_m + m$. We can also pick for $v_{m+1}$ some vector not in the plane spanned by $v_1,\cdots,v_m$, translate $v_1,\cdots,v_m$ by this, and the resulting point $(v_1,\cdots,v_{m+1})$ will still be in $W_{m+1}$. As this modification was linearly independent from the other modifications by definition, we get that $\dim W_{m+1} \geq \dim W_m + m + 1$, or $\operatorname{codim} W_{m+1} < \operatorname{codim} W_m$ and we've proven the result we were after.

  • $\begingroup$ I am happy with the dimension calculation, it is a clever trick. Thanks! However, I do not understand how you can projectivize the map as there are things that map to zero. For example, if the given vectors in $\mathbb{C}^{n\times m}$ are linearly dependent, then there is a linear combination that's mapped to zero. $\endgroup$ – Levent Dec 4 '19 at 13:59
  • 1
    $\begingroup$ This isn't what I'm claiming - I'm not projectivizing the map $\varphi$, I'm projectivizing the sets $\varphi^{-1}(V)$ and $W$. $\endgroup$ – KReiser Dec 4 '19 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.