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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.

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Let $\{i,j,k\} = \{1,2,3\}$ then the simplest thing I can think of is that

$N_i$ and $N_j$ intersect transversely, and $N_k$ transversely intersects $N_i \cap N_j$.

Here's a slightly weaker condition. The statement that two manifolds $N_1, N_2 \subset M$ intersect transversely is equivalent to saying that the product inclusion $N_1 \times N_2 \to M \times M$ transversely intersects the diagonal $\Delta_M = \{ (p,p) : p \in M \} \subset M \times M$.

So you could ask, does the product inclusion $$ N_1 \times N_2 \times N_3 \subset M \times M \times M$$ transversely intersect the diagonal $$\Delta^3_M = \{(p,p,p) : p \in M \} \subset M^3 ? $$

That also gives you the result you're looking for.

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  • $\begingroup$ Thanks, this is certainly a strong enough condition. I guess it's enough if $N_i\pitchfork N_j$ and $N_k\pitchfork(N_i\cap N_j)$ for some (instead of any) triple $i,j,k\in\{1,2,3\}$ of distinct elements. (Perhaps this is what you meant?) I was hoping that some weaker condition would suffice. $\endgroup$
    – Dave
    May 9 '13 at 14:07
  • $\begingroup$ That's a pretty weak condition. Okay, I can think of one weaker condition. I'll write it up. $\endgroup$ May 9 '13 at 15:17
  • $\begingroup$ This condition looks quite nice, thank you very much! $\endgroup$
    – Dave
    May 9 '13 at 17:48
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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 $k_1+k_2+k_3$.

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  • $\begingroup$ This seems like a sensible condition too, thank you. If you have the time to do so, it would be nice if you could expand your answer a bit and add some more details. $\endgroup$
    – Dave
    May 10 '13 at 8:39
  • $\begingroup$ @DaveHartman, see the lengthy edit above. I hope it helps. $\endgroup$ May 10 '13 at 12:11
  • $\begingroup$ Thank you. I like this point of view, it helps a lot. $\endgroup$
    – Dave
    May 10 '13 at 13:06
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    $\begingroup$ In fact this is what I use in everyday life...I am a mechanical engineer and this property is known as non-overconstrainedness... $\endgroup$
    – Troy Woo
    May 29 '15 at 4:01
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You are right that pairwise transverse intersection is not enough; think about three distinct planes in $\mathbb{R}^3$, all containing the $z$-axis.

Ryan's answer is one way to give a sensible condition.

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