# The dimension of the projection of a variety

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.

• 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. – Julian Quast Dec 1 '19 at 20:52
• 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. – 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.

• 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. – Levent Dec 4 '19 at 13:59
• This isn't what I'm claiming - I'm not projectivizing the map $\varphi$, I'm projectivizing the sets $\varphi^{-1}(V)$ and $W$. – KReiser Dec 4 '19 at 17:35