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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.

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  • $\begingroup$ 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 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. $\endgroup$ – user277182 Dec 2 at 6:22
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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.

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  • $\begingroup$ 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 at 13:59
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    $\begingroup$ 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 at 17:35

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