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For a closed connected orientable surface of genus $g$, the Künneth theorem gives the Betti numbers of $S^1\times {\Sigma }_g$ as $(1,2g+1,2g+1,1)$. The weak Morse inequalities give the displayed lower bound. A surface Morse function with one minimum, $2g$ saddles and one maximum exists from its standard handle decomposition. Adding a positive multiple of the cosine Morse function on the circle makes a product of Morse functions, whose critical counts are precisely these four Betti numbers. The lower bound is attained.

For the model $F$ of the Möbius band, the tangent vectors along its core circle of a Möbius band areTheir cross product, divided by its norm $R$, gives the continuous unit normalAlthough $F(2\pi ,0)=F(0,0)$, we have $n(2\pi )=-n(0)$. Any other continuous choice of unit normal along this parametrized interval is either $n$ everywhere or $-n$ everywhere: its relative sign is a continuous map into the discrete space $\{+1,-1\}$, and the interval is connected. Thus no choice can agree at the identified endpoints. This short calculation proves that the Möbius band has no global continuous unit normal and is not an orientable surface.

The plotted quantity is $n(\theta )\cdot n(0)=\cos (\theta /2)$. It compares ambient vectors at generally different points; after one circuit it is $-1$ at the starting point, and after two circuits it is $+1$ there. Only these endpoint comparisons are needed for the orientability argument.

Orientability asks whether local choices of handedness can agree everywhere on a manifold. On a smooth manifold, an orientation of a smooth manifold chooses an orientation of a vector space for each tangent space, continuously with position. An ordered basis describes the same handedness as another one when the change-of-basis determinant is positive. An oriented atlas implements this condition through positive Jacobian determinants of the smooth transition maps.

A paper strip makes the obstruction visible. Join its short ends without twisting: the resulting untwisted band admits a consistent choice of unit normal. Add a half-turn before joining: on the Möbius band, following the core circle of a Möbius band reverses that choice when you first return to the same point. A local decision about handedness has encountered a global contradiction. The normal reversal around the Möbius core makes this precise for an explicit model.

Both rectangles have coordinates $(u,v)\in [0,1]\times [-1,1]$. The untwisted band identifies $(0,v)$ with $(1,v)$; the Möbius band identifies $(0,v)$ with $(1,-v)$. Across a seam the corresponding local coordinate changes preserve or reverse $v$, giving determinants $+1$ and $-1$. This sign explains the handedness change; the drawing also tracks the two boundary circles of the untwisted band and the single boundary circle of the Möbius band.

For a smooth topological surface embedded in ${\mathbb{R}}^3$, choosing an orientation of a surface is equivalent to choosing a continuous unit normal. This picture is specific to surfaces in three-dimensional space: the general definition uses compatible local orientations, without requiring an embedding. Orientability is a topological invariant, so a homeomorphism cannot turn a Möbius band into a untwisted band. The sphere and torus are orientable surfaces; the Klein bottle is not. ^[ref]

The distinction matters when a sign has geometric meaning. A signed flux surface integral needs an orientation of a surface, and Stokes theorem relates that choice to the outward-normal-first boundary orientation. Ordinary unsigned surface area does not require orientability. Thus a Möbius band still has a perfectly meaningful area even though it has no continuous global unit normal.

A connected orientable manifold of positive dimension has two possible orientations: reversing every local choice gives the other one. If a manifold has several connected components, each component can be oriented independently when it is orientable. For topological surfaces, a useful obstruction criterion is the presence of a Möbius band: a surface is nonorientable exactly when it contains such a band. ^[ref]

On the smooth boundary of an orientable surface, let $n$ be the chosen unit normal vector and $t$ the positively oriented unit tangent vector used by Stokes theorem. The outward conormal is $u=t\times n$. It is a unit vector, tangent to the surface and orthogonal to its boundary. This convention turns $(n\times v)\cdot d\mathbf{r}$ into $u\cdot vds$ by cyclic invariance of the scalar triple product.

In Einstein summation convention, the curl of a cross product has componentsThe second line uses the contraction of two Levi-Civita symbols; the last uses the product rule. These components prove the curl of a cross product identity, including both divergence and directional-derivative terms.

Now apply that identity to $m\times v$. On the orientable surface, $m=n$, $m\cdot v=0$, and $|m|=1$. Differentiating $m\cdot m=1$ gives $m_i{\partial }_j{m}_i=0$. Taking a dot product with $m$ in the curl of a cross product formula therefore gives, on $S$,Only tangency on the surface is needed; no claim is made that the extension $v$ is tangent away from it. The right side is the surface divergence of the tangent vector field.

With the boundary oriented according to Stokes theorem, cyclic invariance of the scalar triple product giveswhere $u$ is the outward conormal of a surface boundary. Thus Stokes theorem provesThis is the surface divergence theorem for a tangent field.

For the disc in the $z=0$ plane, take $n=(0,0,1)$ and $v=\mathbf{r}=(x,y,z)$. The field is tangent on the disc, and ${\partial }_j{v}_i={\delta }_{ij}$. The contraction is $3-|n{|}^2=2$, so the surface integral is $2\pi R^2$. On the positively oriented circular boundary, $u=(\cos \theta ,\sin \theta ,0)$ and $v=Ru$, so the boundary flux is $R(2\pi R)=2\pi R^2$. Both sides agree.

Work over $\mathbb{Q}$. A closed connected orientable surface of genus $g$ has Betti numbers $(1,2g,1)$; the circle has Betti numbers $(1,1)$. The Künneth theorem therefore givesWe use the following precise consequence of the Morse handle-attachment theorem: a closed manifold with a Morse function has the homotopy type of a finite CW complex with one index-$j$ cell for each index-$j$ critical point. Its degree-$j$ rational cellular chain group has dimension $c_j$, so its homology has dimension at most $c_j$. Consequently $c_j\ge b_j$, the weak Morse inequalities. Summing yieldsThis counts every critical point, regardless of whether some critical values coincide.

To attain the bound, construct a surface Morse function $h:{\Sigma }_g\to \mathbb{R}$ from the standard orientable handle decomposition: start with a disk, attach $2g$ orientable bands in the usual paired pattern forming the genus-$g$ surface with one boundary component, and cap that boundary with a disk. Give the first disk a minimum, each band one saddle with local form $-x^2+y^2$, and the final cap a maximum. Regular collar coordinates join successive levels without additional critical points. This constructs $h$ with counts $(1,2g,1)$.

On the product takeIts critical points are exactly the pairs of a critical point of $h$ with $\theta =0,\pi$. The Hessian matrix is block diagonal, with surface block $\mathrm{Hess}h$ and circle entry $-\delta \cos \theta$. It is nonsingular and its Morse index is the sum of the two factor indices. Thus this product of Morse functions has countsand exactly $4g+4$ critical points. If distinct critical values are desired, choose $\delta$ outside the finite set producing collisions after first choosing distinct critical values for $h$. The minimal Morse critical count on a circle times an orientable surface is therefore attained for every $g\ge 0$.

For a smooth tangent vector field $v$ on a compact orientable surface with smooth boundary and outward conormal of a surface boundary, the integral of its surface divergence is its boundary flux. Extend the unit normal vector to a unit field $m$. The curl of a cross product gives $m\cdot \nabla \times (m\times v)=\nabla \cdot v-m_i{m}_j{\partial }_j{v}_i$ on the surface: $m\cdot v=0$ there and $m\cdot {\partial }_{j}m=0$ by unit length. Apply Stokes theorem and use $(m\times v)\cdot d\mathbf{r}=v\cdot (d\mathbf{r}\times m)$ to obtain the formula.

A unit normal is a normal vector of Euclidean norm one. Given a nonzero normal vector $n_0$, normalization gives $n=n_0/\Vert n_0\Vert$. This makes a dot product $w\cdot n$ measure the signed length of the normal component, without an extra factor from the length of $n_0$. On a hypersurface, the two opposite choices $n$ and $-n$ correspond to opposite normal orientations; a continuous choice on an orientable surface fixes its orientation of a surface.

An untwisted band is the topological surface obtained from a rectangle by identifying $(0,v)$ with $(1,v)$. It is the Cartesian product of a circle and a closed interval, and is homeomorphic to a closed annulus. A finite portion of the lateral surface of a circular cylinder realizes it in three-dimensional space. Its two boundary circles are $S^1\times \{-1\}$ and $S^1\times \{1\}$, and the matching seam directions make it an orientable surface. The Möbius band changes just one identification sign and loses both this two-component boundary and its orientability.