Sard Theorem and Transversality — Examples

The tangent space at $(x,y,z)\in S^2$ consists of vectors orthogonal to $(x,y,z)$. The differential of $h$ is the projection $v\mapsto v_z$, so it vanishes on that tangent space exactly when every tangent vector has zero $z$-component, which happens only at $(0,0,\pm 1)$.

Therefore the critical locus from [F1] is $\{(0,0,1),(0,0,-1)\}$, and the critical value set is $\{1,-1\}$.

Thus the sphere height function has exactly the two expected critical points and values.

The differential of a constant map is zero at every point. Because $T_{q}N$ is nonzero, this differential is not surjective, so every point of $M$ is critical.

Hence the critical locus from [F1] is all of $M$, and because $M$ is nonempty, its image is the singleton $\{q\}$.

Therefore a constant map has a large critical locus and one critical value.

The supports of the summands are pairwise disjoint, so near each $x$ only finitely many summands are nonzero. Therefore $f$ is smooth. At each center $x=k$, the derivative of the $k$th summand vanishes and every other summand is zero, so $k$ is a critical point with critical value $f(k)=1-\frac{1}{k}$.

The sequence $1-\frac{1}{k}$ tends to $1$. But $f(x)\mathrm{\ne }1$ for every $x$: on each bump support the value is at most $1-\frac{1}{k}<1$, and away from the supports the value is $0$. Thus $1$ has empty fibre and is therefore regular.

By [F1], the critical value set contains every $1-\frac{1}{k}$ but not the limit $1$, so it is not closed. This is exactly the phenomenon noted in [L1].

The tangent lines to $L_1$ and $P$ are spanned by $(1,1)$ and $(1,2x)$, respectively. At $x=0$ and $x=1$ these are distinct, so their spans add to ${\mathbb{R}}^2$. Thus $L_1⋔P$ at both intersection points by [F1].

The tangent line to $L_0$ is spanned by $(1,0)$, and the tangent line to $P$ at $(0,0)$ is also spanned by $(1,0)$. Their sum is only one-dimensional, so [F1] fails there. This is the tangent situation warned about by [L1].

Therefore the parabola exhibits both transverse and tangent intersections in the plane.

If $F(x)=(1,1)$, then $(x_1,x_2)\mathrm{\ne }(0,0)$ and $(x_3,x_4)\mathrm{\ne }(0,0)$. The Jacobian matrix $\left[\begin{array}{cccc}2x_1& 2x_2& 0& 0\\ 0& 0& 2x_3& 2x_4\end{array}\right]$ therefore has rank $2$, so $(1,1)$ is a regular value.

The fibre equation is exactly $x_1^2+x_2^2=1,x_3^2+x_4^2=1,$ which is $S^1\times S^1$. By [L1], this fibre is an embedded codimension-$2$ submanifold of ${\mathbb{R}}^4$.

Thus the product of the two coordinate circles appears as a transverse level set.

If $F(x,y)=G(u,v)$, then both differentials are the row matrix [L1, given, algebra] $\left[\begin{array}{cc}1& 0\end{array}\right]$, so $d{F}_{(x,y)}(T_{(x,y)}{\mathbb{R}}^2)+d{G}_{(u,v)}(T_{(u,v)}{\mathbb{R}}^2)=\mathbb{R}\mathrm{.}$ Therefore [L1] gives $F⋔G$.

By [L2], the fibre product is an embedded submanifold of ${\mathbb{R}}^4$. [L2, step 1.1, algebra] Its defining equation $x=u$ cuts the ambient dimension down by one, so it is $3$-dimensional.

Thus the coincidence set of two coordinate projections is a concrete fibre product of submersions.

The fibre $h^{-1}(a)$ is $S^1\cap H_a$. When $\mathrm{\mid }a\mathrm{\mid }<1$, the equation $x=a$ on $x^2+y^2=1$ gives the two points $(a,\pm \sqrt{1-a^2})$. At either point, $dh$ is nonzero on the circle tangent line exactly when $a\mathrm{\ne }\pm 1$.

Therefore for every $a\mathrm{\ne }\pm 1$, the line $H_a$ either misses the circle or meets it transversely. The exceptional set $\{-1,1\}$ is finite, hence null, exactly as [L1] predicts in this one-parameter family.

Thus generic affine hyperplanes meet the embedded circle transversely.

The image of $F$ is the single point $(0,1)$, which does not lie on the $x$-axis. Hence $F^{-1}(Z)=\mathrm{\varnothing }$.

By [F1], the transversality condition has no points to check when the preimage is empty. Therefore $F⋔Z$.

This is a concrete vacuous transversality example.

The intersection $S\cap T$ is the whole $x$-axis, so it is $1$-dimensional.

Each of $S$ and $T$ has codimension $1$ in ${\mathbb{R}}^2$, so a transverse intersection would have expected codimension $2$ and thus expected dimension $0$. Step 1.1 shows the actual intersection dimension is larger. This is precisely the nontransverse situation highlighted in [L1].

Therefore tangent intersections need not have the expected set-theoretic dimension.