Let $X$ and $Z$ be transversal submanifolds of $Y$. Prove that if $y \in X \cap Z$, then $$T_y(X \cap Z) = T_y(X) \cap T_y(Z).$$

("The tangent space to the intersection is the intersection of the tangent spaces.")

Given $X$ and $Z$ transversal submanifolds of $Y$, we know $$T_x(X) + T_x(Z) = T_x(Y)$$

So my idea is first show $\forall v \in T_y(X \cap Z), v \in T_y(X) \cap T_y(Z)$.

So if $v \in T_y(X \cap Z)$, it is in the tangent plane of the intersection of $X$ and $Z$, so can I draw the conclusion that it is in the tangent plane of $X$ and $Z$ directly from here?

And then the other direction, if $ v \in T_y(X) \cap T_y(Z)$, then it is in the tangent plane of $X$ and $Y$ respectively, then can I just say it is also in $T_y(X \cap Z)$?

Please kindly advice - thanks!!


First, note that since $X \cap Z \subseteq X$ and $X \cap Z \subseteq Z$, we have that for $y \in X \cap Z$, $T_y(X \cap Z) \subseteq T_y(X)$ and $T_y(X \cap Z) \subseteq T_y(Z)$, or $$T_y(X \cap Z) \subseteq T_y(X) \cap T_y(Z),$$ by the definition of intersection.

Since $X \pitchfork Z$, we have that $$\mathrm{codim}(X \cap Z) = \mathrm{codim}(X) + \mathrm{codim}(Z),$$ or $$\dim(X \cap Z) = \dim(X) + \dim(Z) - n,$$ which tells us that $$\dim(T_y(X \cap Z)) = \dim(T_y(X)) + \dim(T_y(Z)) - n.$$ On the other hand, from linear algebra we know the formula for the dimension of an intersection of vector spaces: \begin{align*} \dim(T_y(X) \cap T_y(Z)) & = \dim(T_y(X)) + \dim(T_y(Z)) - \dim(T_y(X) + T_y(Z)) \\ & = \dim(T_y(X)) + \dim(T_y(Z)) - n, \end{align*} where $\dim(T_y(X) + T_y(Z)) = n$ since $X \pitchfork Z$. Then since $T_y(X \cap Z) \subseteq T_y(X) \cap T_y(Z)$ and both vector spaces have the same dimension, it must be that they are equal. We conclude that for all $y \in X \cap Z$, $$T_y(X \cap Z) = T_y(X) \cap T_y(Z).$$

Note that we can't automatically conclude the second part. Here transversality is important. For example if $$X = \{(x, 0) : x \in \Bbb R\} \subset \Bbb R^2, \quad Z = \{(x,x^2) : x \in \Bbb R\} \subset \Bbb R^2,$$ we have that $X \cap Z = \{(0,0)\}$, so that $T_{(0,0)}(X \cap Z) = \{0\}$, a zero-dimensional vector space. But $T_{(0,0)} (X) = T_{(0,0)} (Z)$ is a $1$-dimensional vector space, so that we necessarily have a strict inclusion $$T_{(0,0)}(X \cap Z) \subset T_{(0,0)}(X) \cap T_{(0,0)} (Z).$$ The problem here is that $X$ and $Z$ do not intersect transversally.

  • $\begingroup$ Beautiful, thanks so much Henry! I really like how you prove the other direction not reversely but from the dimension. And I am very thankful for the counter example helping me clear off my wrong idea! $\endgroup$ – 1LiterTears Jul 9 '13 at 2:28

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