1
$\begingroup$

In Audin and Damian, p.44-45, there seems to be a claim that one can prove transversality without directly showing the tangent spaces span the tangent space of the ambient manifold. In particular, in the following enter image description here

the last sentence seems to claim that if two manifolds intersect in a submanifold, and the codimension of the intersection in the first equals the codimension of the second in the ambient manifold then they are transverse.

This would contradict a line tangent to a parabola in the plane, however. Would someone understand how transversality is implied here?

$\endgroup$
1
  • 1
    $\begingroup$ Aren’t you forgetting the hypothesis that $w$ is a regular value of $g$? $\endgroup$ Commented Jul 24, 2021 at 3:46

2 Answers 2

1
$\begingroup$

I think I understand the issue now. Let me call the submanifolds of $V^n$ by $W^{n-k}$ and $P^{n-l}$. Fix a point in $p$ in $W\cap P$.

The condition $codim_P W^{n-k}\cap P^{n-l} = codim_V W^{n-k}$ implies that $W^{n-k}\cap P^{n-l}$ has dimension $n-k-l$.

In particular, the largest linearly independent set in $T_p(W\cap P)$ is of size $n-k-l$, say $$v_1,...,v_{n-k-l}.$$ Since $T_p W$ has dimension $n-k$, we may choose $l$ more linearly independent tangent vectors $$u_{n-k-l+1},...,u_{n-k}$$ which complete a basis in $T_pW$, but none of these lie in $T_p (W\cap P)$.

There are also $k$ tangent vectors $$w_{n-k+1},...,w_n$$ in $T_pP$ which complete a basis for $T_pP$ with the $v_i$.

Claim: These $v_i$'s, $u_i$'s, and $w_i$'s together form a basis for $T_pV$, and we can do this at every point, showing transversality.$\square$

Justification: In fact, we can start with the linearly independent set $v_1,...,v_{n-k-l},u_{n-k-l+1},...,u_{n-k}$ in $T_pV$, and start filling it with the $w_i$.

If some $w_j$ is a linear combination of the previous ones, such linear combination can be chosen not to include the previous $w$'s (say, by subtracting whatever contributions they had first, and keeping the label $w_j$ for it). Then looking at the tangent space as the space of germs of curves at $p$, $$w_j = \sum_{i=1}^{n-k-l} \lambda_iv_i + \sum_{i=n-k-l+1}^{n-k} \mu_iu_i \Rightarrow w_j - \sum_{i=1}^{n-k-l} \lambda_iv_i = \sum_{i=n-k-l+1}^{n-k} \mu_iu_i.$$

The LHS says there's a regular curve centered at $p$ in $P$ that generates it. The RHS says there's a regular curve centered at $p$ in $W$ that generates it.

This means both curves agree on a small neighborhood $(-\epsilon,\epsilon)$. This means this piece of curve (the restrictions of both to $(-\epsilon,\epsilon)$) lie in both $W$ and $P$, and hence the underlying tangent vector must be a linear combination of the $v_i$'s, say $\sum\nu_iv_i$.

Then $w_j$ becomes $\sum \nu_iv_i + \sum \lambda_iv_i$, a contradiction.

$\endgroup$
1
$\begingroup$

The fact that $W^s_{X'} \cap P$ is a codimension $k$ submanifold of $P$ follows from the regular value theorem. The tranversality follows from Grassmann's formula. If $x \in W^s_{X'} \cap P$, then we have $$ \dim (T_x W^s_{X'} + T_x P) = \dim (T_x W^s_{X'}) + \dim (T_xP) - \dim \big(T_x (W^s_{X'} \cap P) \big) = n - k + k = n. $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .