The current sign for transverse three-tori in a six-torus

Put $N=(0,1)$, $S=(0,-1)$, $U_N=S^1\setminus \{N\}$ and $U_S=S^1\setminus \{S\}$. The maps ${\sigma }_N(x,y)=x\mathrm{/}(1-y)$ and ${\sigma }_S(x,y)=x\mathrm{/}(1+y)$ have inverses ${\sigma }_N^{-1}(u)=(\frac{2u}{1+u^2},\frac{u^2-1}{1+u^2}),{\sigma }_S^{-1}(u)=(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2})\mathrm{.}$ Substitution verifies both inverse identities; their rational components are smooth. The subspace $S^1\subset {\mathbb{R}}^2$ is Hausdorff, and intersections with Euclidean balls having rational centers and positive rational radii give a countable base. The sets $U_N,U_S$ are open and cover $S^1$; the displayed maps and inverses are continuous, so they are homeomorphisms onto $\mathbb{R}$ and exhibit a topological $1$-manifold. On their overlap both transition maps are $u\mapsto 1\mathrm{/}u$, smooth on $\mathbb{R}\setminus \{0\}$. Thus they give a smooth atlas, whose maximal smooth structure makes $S^1$ a smooth manifold by [F4]. In each chart the inverse parametrization has derivative norm $2\mathrm{/}(1+u^2)>0$, so it is an immersion; as the chart inverse onto the corresponding open arc it is also a topological embedding. Hence the inclusion $S^1\hookrightarrow {\mathbb{R}}^2$ is a smooth embedding under [F5]. The vector field $(-y,x)$ is tangent because its dot product with $(x,y)$ is zero, and nonzero because $x^2+y^2=1$; its smooth components define the stated orientation under [F6].

The unit circle is closed and bounded in ${\mathbb{R}}^2$, so it is compact by [F7]. A singleton is compact, and [F8] makes $X,C,D$ compact. Product charts from [F3] give $X$ its product smooth structure and make $C,D$ local coordinate slices; their inclusions are smooth embeddings by [F5]. By the explicit factor-order orientation in the Example, the first three positive factor tangents orient $C$, the last three orient $D$, and their concatenation is positive for $X$. The defining coordinate conditions show $C\cap D=\{(y_1,\dots ,y_6)\}$. Their tangent spaces are the spans of the first and last three factor tangents, so they sum directly to $T_{x}X$ and meet transversely. The C-first local intersection sign is therefore $+1$.

Here $p(n-p)=3\cdot 3=9$, so [F1] gives $C{\cdot }_{\mathrm{c}\mathrm{u}\mathrm{r}}D=(-1{)}^9(+1)=-1$. [F2] identifies that number with $⟨{\delta }_C,D{⟩}_{\mathrm{r}\mathrm{e}\mathrm{g}}$.