TITLE: Transverse intersection of multiple submanifolds
QUESTION [5 upvotes]: Let $M$ be a smooth manifold and suppose that we have three (or more) submanifolds $N_1,N_2,N_3\subset M$.
What is the right notion of "transverse intersection" of $N_1,N_2,N_3$, i.e. what is the (weakest) condition such that 
$$N_1\cap N_2\cap N_3\subset M
$$
a submanifold of codimension
$$codim (N_1\cap N_2\cap N_3)=\sum_{i=1}^3codim N_i \quad?
$$
I think that pairwise transverse intersection 
$$\forall i\neq j: N_i\pitchfork N_j
$$
is not enough.
Relevant references are also appreciated.

REPLY [5 votes]: For obvious reasons, in what follows I'm going to rename the submanifolds $X_i$. I claim that transversality can be rephrased in terms of the normal bundles $N_i = N(X_i,M)$ of $X_i$ in $M$.
Start with the case of two submanifolds $X_1$ and $X_2$ of codimensions $k_1$ and $k_2$. Locally (on $U\subset M$), they are given as the zero-set of functions $f_i\colon U\to\mathbb R^{k_i}$ that have maximum rank. To say that $X_1\pitchfork X_2$ is to say that the function $f=(f_1,f_2)\colon U\to\mathbb R^{k_1+k_2}$ has maximum rank. (This follows easily from the nullity-rank theorem.) Now, geometrically, the normal space of $X_i$ is spanned by the row vectors of $Df_i$ (i.e., the gradient vectors of the components). To say that $X_1\pitchfork X_2$ is to say that the gradient vectors of all the component functions of $f$ are linearly independent. That is, $N(X_1\cap X_2,M) = N(X_1,M)\oplus N(X_2,M)$.
This viewpoint generalizes to any number of submanifolds:  $X_1\cap X_2\cap X_3$ will be a submanifold when $N_x(X_i,M)$ are linearly independent in $T_xM$ for every $x\in X_1\cap X_2\cap X_3$. In this event, $N(X_1\cap X_2\cap X_3,M) = N(X_1,M)\oplus N(X_2,M)\oplus N(X_3,M)$ has rank $k_1+k_2+k_3$.