# Understanding transverse

I have read "let $f:X \rightarrow Y$ be a $C^{\infty}$ manifold without a boundary and let $Z \subset Y$ be a submanifold of $Y$. Geometrically, we say that $f$ is transverse to $Z$ if $X \times Z$ and $Gr(f) = \{(x, y) \in X \times Y |f(x) = y \}$ are in general position." What does it mean?

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Wherever you read the above expression, did they earlier define general position? Usually transversality means that if $x \in f^{-1}(Z)$ then the image of $Df_x : T_x X \to T_{f(x)} Y$ together with $T_{f(x)}Z$ span $T_{f(x)} Y$. – Ryan Budney Mar 27 '11 at 17:43

## 1 Answer

Although this certainly isn't quite right, here is what I visualize in my mind when I hear the word transverse. The map $f$ has a differential $df_x$ at every point. If we only consider points on $X$ that map into $Z$, say $p\in f^{-1}(q)$, then $df_p: T_pX\to T_qY$.

In some sense $\mathrm{Im}(df_p)$ is measuring the "direction" of $f$ in $Y$. $Z$ also has a "direction" in $Y$ and we could think of this as $T_qZ\subset T_qY$. We say $f$ is transverse to $Z$ if the collection of all (span of) these directions is the full tangent space at $q$, or $\mathrm{Span}(im(df_p), T_qZ)=T_qY$

For example, take the map $f: \mathbb{R}\to \mathbb{R}^2$ given by $f(x)=(x,0)$. Take as a submanifold of $\mathbb{R}^2$ the $y$-axis. Then the only place where the image of $f$ (which is the $x$-axis) intersects the submanifold is at $(0,0)$. So we only have one point to check, but $im(df_0)=\frac{\partial}{\partial x}$ and $T_{(0,0)}Z=\frac{\partial}{\partial y}$, so the span of these two is all of $T_{(0,0)}\mathbb{R}^2$.

What we see here is that the image of $f$ is the $x$-axis and it intersects the submanifold (the $y$-axis) in a nice way. Note that the $y$-axis could have been any something like the line $y=x$ and it still would be transverse since we are taking span. It didn't need to be perpendicular. If the submanifold was $y=x^2$ then since the intersection point is tangent, we get an example of something that is not transverse.

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Thank you. Would you kindly explain them using an example of some functions? – Thales Mar 27 '11 at 17:08
Hopefully I now start to get the idea. Like that these two directional vectors are not dependent (sorry for a primitive expression)? $T_{0,0}$ is a tangent plane at $(0,0)$. What is $Im(df_p)$? Thank you so much. – Thales Mar 27 '11 at 17:26
$df_p$ is just a linear transformation between the vector spaces $T_0\to T_{0,0}$, so $Im$ just means the image of the vector space under that linear transformation. – Matt Mar 28 '11 at 1:39
Thales: your use of independent is precisely the point if by independent you mean linearly-independent. Notice that in Matt's example, the two vectors; the one transported from the domain manifold (using the pushforward) map, and the vector at the tangent space of the (embedded) target Z, span the entire tangent space of the ambient space M, because the two vectors are linearly independent. These two vectors , being linearizations, carry some information on how the image of f intersects Z. – gary Apr 4 '11 at 6:07
Another test for transversality of submanifolds M,N of X, is that the dimension of the intersection of M and N in X equals the dimension of X minus the sum of the respective codimensions of M,N, i.e., Dim(M/\N)=DimX-Codim(N)-codim(M) – gary Apr 4 '11 at 6:25