Let $\hbox{Aff(}k,n)$ be the space of $k$-dimensional affine subspaces of $R^n$. The group of Euclidean isometries of $R^n$ (the semi-direct product of rotations and translation) acts transitively on $\hbox{Aff(}k,n)$, so by general Lie theory $\hbox{Aff(}k,n)$ is a smooth manifold and there is a smooth invariant measure on it (called kinematic measure in Integral Geometry) that is unique to within positive scalar multiple. If $M$ is a smooth $(n-m)$-dimensional submanifold of $R^n$, then it is an easy corollary of the Parametric Transversality Theorem that all $F \in \hbox{Aff(}k,n)$ outside a set $E$ of measure zero are transversal to $M$, so that if $F \notin E$ then $F\cap M$ is a smooth $(n-(m+k))$-dimensional submanifold of $R^n$. (Special case: Almost all lines in $R^3$ intersect a given surface in a discrete set.) Of course this has to be a well-known theorem, and I would like to know the appropriate reference for it.

Dear Richard,

I think this is just thought of as "folklore" in integral geometry (sorry, never seen it written down). This is probably because the statement you give follows from something that needs almost no technology to prove:

The set of $k$-flats that are *parallel* to a given $k$-dimensional subspace $L$ can be identified $R^n / L$ and provided with the Lebesgue measure. Among these flats, those that intersect the manifold $M$ transversely have full measure in $R^n / L$.

One then uses the usual decomposition of the kinematical measure (the set of affine $k$-flats is the total space of a tautological bundle over a Grassmannian) to conclude the statement you seek.

This is a special case of a more general theorem on generic transversality of submanifolds in homogeneous spaces:

Let $Y$ be a smooth homogeneous space for a Lie group $G$ and $Z_1, Z_2\subset Y$ be smooth submanifolds. Then for any measure in the Lebesgue measure class on $G$, for almost all $g\in G$ the intersection $Z_1\cap g Z_2$ is transversal.

Algebraic geometers know this result as *Kleiman Transversality Theorem* (in the context of smooth algebraic varieties and algebraic group actions), see

S.L. Kleiman, The transversality of a general translate, Compositio Math. 28 (1974), 287–297.

When we needed this (actually, a bit more general) result in the category of smooth manifolds and actions we could not find this (folklore) theorem in the literature, so we proved it in Proposition 2.3 of M. Kapovich, B. Leeb and J. Millson, "Convex functions on symmetric spaces, side lengths of polygons and the stability inequalities for weighted configurations at infinity," Journal of Differential Geometry, Vol. 81, 2009, p. 297-354.

The result that we proved is slightly more general and deals with several smooth submanifolds of $Y$ and their $G$-translates.

Assuming that by "smooth measure" you mean a measure coming from a smooth top-dimensional form, we don't need a particular measure to talk about sets of measure zero on manifolds: $E$ is of measure zero for one such measure if and only if it is of measure zero for any other.

Consider the space $S=\mathbb{R}^n \times V_k(\mathbb{R}^n) = \{(p, e_1, e_2, \ldots, e_k) | p \in \mathbb{R}^n,
(e_i)_{i=1}^k$ is an an orthonormal system of vectors in $\mathbb{R}^n\}$, the product of $\mathbb{R}^n$ and the *Stiefel manifold* of $\mathbb{R}^n$ (I like to think about it as a bundle of Stiefel manifolds on tangent spaces). This is also a manifold, and this is a (smooth fiber) bundle over $\mathrm{Aff}(k,n)$: to each such tuple $s=(p,e_1,\ldots,e_k)$ one assigns the affine space $\Pi(s)$ spanned by $e_i$ at $p$, parametrized by $f_s(x_1, \ldots, x_k) = p+\sum_{i=1}^k x_i e_i$. We denote the map by $\Pi:S \to \mathrm{Aff}(k,n)$.

Thom Parametric Transversality Theorem now says that for almost all elements $s \in S$, the map $f_s$ is transversal to $M \subset \mathbb{R}^n$.

Suppose now the set $E$ of all $k$-dimensional affine spaces which do not intersect $M$ transversally has non-zero measure. Then its preimage $\Pi^{-1}(E) \subset S$ has non-zero measure as well (since $\Pi$ is a smooth *submersion*). But the transversality of $f_s$ to $M$ is equivalent to the transversality of $\Pi(s)$ to $M$, so $f_s$ is not transversal for each $s \in \Pi^{-1}(E)$. Which contradicts the Thom Parametric Transversality Theorem.