Let $V$ be a analytic/algebraic variety in $\mathbb{CP}^n$, that is $V$ is the vanishing locus of some homogeneous polynomial. Let $d = \dim V$. Everyone knows$^\mathrm{TM}$ that a "generic $k$-plane" $H \subset \mathbb{CP}^n$ intersects $V$ transversely. "Generic" here means that this is true for all $H \in \mathrm{Gr}(k+1, n+1)$ that don't lie in a proper subvariety of $\mathrm{Gr}(k+1, n+1)$.

I have two questions / reference requests:

First, what do people mean by "transverse" when $V$ is singular? Do you inductively define $H \pitchfork V$ iff $H \pitchfork V_\mathrm{reg}$ an $H \pitchfork V_\mathrm{sing}$?

Second, how do you actually prove that this is generic in the Zariski-sense?


My thoughts:

Using some kind of parametric transversality theorem (e.g. Lee, Introduction to smooth manifolds Thm. 6.35) you can show that this is true for a full measure set of $\mathrm{Gr}(k+1, n+1)$ but this is quite far from showing it's Zariski-open.

The following approach seems promising but I can't get it to work: Consider a parametrization of embeddings of $k$-planes into $\mathbb{CP}^n$ by some complex manifold $S$, i.e. we have a map $F: S \times \mathbb{CP}^k \to \mathbb{CP}^n$ s.t. the images of $\{s\} \times \mathbb{CP}^k$ are all possible $k$-planes in $\mathbb{CP}^n$. If $F$ restricted to $F^{-1}(V)$ is proper then I think the proof of Sard's theorem/parametric transversality together with Remmert's proper mapping theorem give you the claim. However I don't see how to construct a $F$ like this that is proper.

  • $\begingroup$ You want to look up Bertini's theorem. I'm a little busy right now but I'll see if I can come back and write a few answer in the next couple days. $\endgroup$
    – KReiser
    Nov 5, 2021 at 7:19

1 Answer 1


There are some different definitions of transverse intersection in the literature. We'll go through some of the tradeoffs for your problem and then explain the method of proof. Suppose $X_1$ and $X_2$ are closed subvarieties of some variety $X$.

  • Every definition I've seen requires $T_xX_1+T_xX_2=T_xX$ for $x\in X_1\cap X_2$ in order to say that $X_1$ and $X_2$ intersect transversely at $x$.
  • Many definitions also require that $X_1$, $X_2,$ and $X$ are all nonsingular at $x$ in order to say that $X_1$ and $X_2$ intersect transversely at $x$.

Definition #2 leads to conflict with your statement that a generic $k$-plane intersects $X$ transversely: if we take a variety that has a singular locus of dimension $d$ and consider planes of dimension $n-d$ or more, we're always going to have an intersection between the plane and the singular locus for dimension reasons. Therefore to use condition #2 and prove your statement, one needs to add an assumption about the dimension of the singular points. We'll ignore the condition and prove the claim without this.

The strategy is to consider the incidence correspondence $$\mathcal{I}=\{ (W,p) \mid W \text{ intersects } V \text{ non-transversely at } p\}\subset G(m+1,n+1)\times V.$$ We'll show that it's a union of finitely many locally closed subvarieties, each of which has dimension less than $\dim G(m+1,n+1)$ so that when we project to $G(m+1,n+1)$, all the $m$-planes in the complement of the image intersect $V$ transversely everywhere.

We'll start by showing that $\mathcal{I}$ is a union of locally closed subvarieties. Choose $f_1,\cdots,f_r$ which generate the ideal of $V$. Then the dimension of the tangent space at a point $p\in\Bbb P^n$ is $n$ minus the rank of the projective Jacobian $\{\frac{\partial f_i}{\partial x_j}\}_{1\leq i\leq r, 0\leq j\leq n}$ evaluated at $p$. Now choose $n-m$ linearly independent equations $\sum a_{ij}x_j$ which cut out our $m$-plane $W$: the condition that $T_pV+T_pW=T_p\Bbb P^n$ can be expressed in terms of the rank  of the Jacobian matrix with $\{a_{ij}\}$ appended to the end. Specifically, $n$ minus the rank of the Jacobian with the linear equations appended will be the dimension of the intersection of the tangent space with $W$. Therefore we can form the set $\mathcal{I}_i = \{(W,p)\mid \dim T_pV = i, \dim T_pV\cap T_pW > i+m-n\}$ which is locally closed in $G(m+1,n+1)\times V$ and the union of these over all $i$ is $\mathcal{I}$.

To show the claim about dimensions, let's find the dimension of $\mathcal{I}_i$ for each $i$. In order to bound the dimension of $\mathcal{I}_i$, it suffices to analyze the dimension of the fiber of the projection $\mathcal{I}_i\to V$: the dimension of the incidence correspondence is at most the dimension of the base plus the dimension of the fiber.

To find the dimension of the fiber, we need to answer the following question: if we have an $n$-dimensional vector space with a fixed subspace of dimension $k$, what's the dimension of the $m$-dimensional subspaces which have at least an $\ell$-dimensional intersection with our fixed subspace? Up to a coordinate change, we may assume our fixed subspace is the span of the first $k$ basis vectors, which means we can find the answer in the following matrix:

$$\begin{pmatrix} I_\ell & \ast & 0 & 0\\ 0 & I_{k-\ell} & 0 & \ast \\ 0 & 0 & I_{m-k-\ell} & \ast \end{pmatrix}$$

This represents an $m$-dimensional subspace which has intersection of dimension at least $k$ with the fixed subspace. The dimension is just the number of asterisks: $\ell(k-\ell)+(m-\ell)(n-m)$. Substituting $\ell=i+m-n+1$ and $k=i$ above, we find that the dimension of the non-transverse hyperplanes in the fiber $G(m,n)$ over a point $v\in V$ is $m(n-m)-i-m+n-1=(m+1)(n-m)-i-1$.

We can note that $\mathcal{I}_i$ is empty for $i<\dim V$, as the tangent space is always of dimension at least $\dim V$. When $i=\dim V$, the projection of $\mathcal{I}_i$ to $V$ is exactly the collection of nonsingular points of $V$, and it's a dense open set of $V$, while when $i>\dim V$, the projection of $\mathcal{I}_i$ to $V$ lies in the singular locus, a proper subvariety (and therefore of dimension strictly less than $\dim V$).

So when $i=\dim V$, we find $\dim \mathcal{I}_i$ is at most $(m+1)(n-m)-i-1+i$, which is less than $(m+1)(n-m)$, and when $i>\dim V$ we see that $\dim \mathcal{I}$ is at most $(m+1)(n-m)-i-1+\dim V-1$, which is again less than $(m+1)(n-m)$. Thus $\dim \mathcal{I} < \dim G(m+1,n+1)$ and therefore the projection of $\mathcal{I}$ to $G(m+1,n+1)$ is contained in a proper subvariety, so every $m$-plane in the complement intersects $V$ transversely.

  • $\begingroup$ I am familiar with the idea of this argument. The part that is not quite clear to me is why the projection of $\mathcal{I}$ to $G(m+1, n+1)$ is known to be a variety. I guess that maybe isn't necessary if we know that the $\mathcal{I}_i$ are locally closed, because then the images of their Zariski-closures are subvarieties (of positive codimension) of $G(m+1, n+1)$ by the proper mapping theorem. Can you elaborate on how to show that the $\mathcal{I}_i$ are locally closed? $\endgroup$ Nov 15, 2021 at 18:33
  • $\begingroup$ Your sentence starting "I guess..." is correct. As far as showing that the $\mathcal{I}_i$ are locally closed, can you say a little more about where you'd like additional details? The key idea is that the condition "the rank of a matrix is less than $n$" is determined by the vanishing of the determinant of all $(n-1)\times (n-1)$ minors, and these are polynomials in the entries and so this is a closed condition. From there you can say that the condition "the rank of a matrix is at least $n$" is open, and so conditions about the rank are locally closed. $\endgroup$
    – KReiser
    Nov 15, 2021 at 19:50
  • $\begingroup$ So is the idea for showing that $\mathcal{I}_i$ are locally the closed this? $\mathcal{I}_i$ is the intersection of three conditions (globally defined on $V \times G(\dots)$): $\dim T_p V \geq i$ (closed), $\dim T_p V < i+1$ (open) and the third one. If we say $f_1, \dots, f_r$ are (local) defining functions for $V$ then the third condition is locally equivalent to $\dim (d_p f_1, \dots, d_p f_r)(T_p W) < r + n - m - i$ which is closed w.r.t. $p$ and $W$. So the third condition is closed around every point of $V \times G(\dots)$ so it is a closed condition. $\endgroup$ Nov 15, 2021 at 20:01
  • $\begingroup$ Yes, that's pretty much right - it's important to be able to say that the third condition is global, though, and one can actually do that the same way that one globalizes the first condition. $\endgroup$
    – KReiser
    Nov 15, 2021 at 22:22
  • 1
    $\begingroup$ 1) It's not quite that simple. Consider a curve with a single node in the plane and one of the tangents to the node: that intersection is transverse at the node in the sense of my answer because the tangent space at the node is 2-dimensional. But no line is transverse to any reduced closed point in $\Bbb P^2$. 2) For global, either of those two things you state in your final comment is fine - I said it because it looked like you were evaluating at $P$ and attempting to conclude closedness from closedness on each fiber. $\endgroup$
    – KReiser
    Nov 16, 2021 at 2:29

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