Given two differentiable manifolds $\mathcal M$ and $\mathcal N$ I needed to show that $$\mathcal M, \; \mathcal N \mbox{ orientable } \Rightarrow \mathcal M \times \mathcal N \mbox{ orientable.}$$ This was posed as exercise 4.3-3 on p. 141 in Marsden's 'Introduction to Mechanics and Symmetry' book.
I already have a solution (I think), but I'm not really happy with it because it uses coordinates and it's way too long for my taste.
So my questions concern the solution that I have included below. They are:
- Is it possible to make my proof completely coordinate independent? (This concerns Lemma 1)
- Can the proof be made shorter somehow? (The fact that the proved statement seems rather trivial and the proof turned out rather long is really bewildering to me.)
- Is there a better way to prove Lemma 3? Maybe using derivations instead?
- Are there any mistakes?
Thanks in advance for any suggestions.
I know that there have already been several questions / discussions pertaining to this particular problem on here but they are either using concepts from algebraic topology that I don't understand (yet?), or work wrt. to bases, or in charts (using jacobian determinants) both of which I try to avoid whenever possible. The most concise answer on here I have found in this post which practically agrees with my solution attempt (and even shows the converse statement) but for my taste still works too much wrt. to coordinates.
Notation:
- $\oplus$ denotes the inner direct sum of vector spaces, i.e. let $C$ be a vector space, and $A \subset C, B \subset C$ linear subspaces then $A \oplus B = C$ means $C \subset A + B$ and $A \cap B = \{0\}$.
- The tangent map for a map $\varphi: \mathcal M \to \mathcal N$ between differentiable manifolds $\mathcal M$ and $\mathcal N$ is denoted by $T\varphi: \mathcal{TM} \to \mathcal {TN}$ (notation as used in the book).
- The pullback mapping for a map $\varphi: \mathcal M \to \mathcal N$ between differentiable manifolds $\mathcal M$ and $\mathcal N$ is denoted by $\varphi^*$.
My solution attempt:
Theorem. $\mathcal M, \; \mathcal N \mbox{ orientable } \Rightarrow \mathcal M \times \mathcal N \mbox{ orientable.}$
$\mathcal M, \mathcal N$ orientable implies that there exist nowhere vanishing volume forms $\mu$ and $\nu$ on $\mathcal M$ and $\mathcal N$, respectively. Let $\pi_{\mathcal M}: \mathcal M \times \mathcal N \to \mathcal M,\; \pi_{\mathcal N}: \mathcal M \times \mathcal N \to \mathcal N$ be the natural projection maps. Then by Lemma 1 $$\omega = \pi_{\mathcal M}^* \mu \wedge \pi_{\mathcal N}^* \nu,$$ is a nowhere vanishing volume form on $\mathcal M \times \mathcal N$. Hence $\mathcal M \times \mathcal N$ is orientable.$\square$
Lemma 1. $\omega$ is a nowhere vanishing volume form.
Let $\{\mathrm d a^i\},\; \{\mathrm d b^i\}$ be local bases for $\mathcal T^* \mathcal M$ and $\mathcal T^* \mathcal N$, respectively. Then by Lemma 2 $\{dx^i:= \pi_{\mathcal M}^*\mathrm d a^i\} \cup \{dy^i:= \pi_{\mathcal N}^*\mathrm d b^i\}$ is a local basis for $\mathcal T^*(\mathcal M \times \mathcal N).$
Wrt. these local bases one has$$ \begin{align} \mu &= \mu' \mathrm d a^1 \wedge \cdots \wedge \mathrm d a^m,\\ \nu &= \nu' \mathrm d b^1 \wedge \cdots \wedge \mathrm d b^n,\\ \end{align} $$where $m = \mathrm {dim} \mathcal M,\; n = \mathrm {dim} \mathcal N,$ and $\mu',\; \nu'$ are nowhere vanishing smooth functions on $\mathcal M$ and $\mathcal N$, respectively. Then $$ \begin{align} \omega &= \pi_{\mathcal M}^* \mu \wedge \pi_{\mathcal N}^* \nu\\ &= (\pi_{\mathcal M}^* \mu' \pi_{\mathcal M}^* \mathrm d a^1 \wedge \cdots \wedge \pi_{\mathcal M}^* \mathrm d a^m) \wedge (\pi_{\mathcal N}^* \nu' \pi_{\mathcal N}^* \mathrm d b^1 \wedge \cdots \wedge \pi_{\mathcal N}^* \mathrm d b^n)\\ &= \mu' \circ \pi_{\mathcal M} \, \nu' \circ \pi_{\mathcal N} \, \mathrm d x^1 \wedge \cdots \wedge \mathrm d x^m \wedge \mathrm d y^1 \wedge \cdots \wedge \mathrm d y^n \end{align}$$ is clearly a nowhere vanishing volume form on $\mathcal M \times \mathcal N.\,\square$
Lemma 2. $\mathcal T^*(\mathcal M \times N) = \pi_{\mathcal M}^* (\mathcal T^* \mathcal M) \oplus \pi_{\mathcal N}^* (\mathcal T^* \mathcal N).$
First observe that the pullback wrt. to $\pi_{\mathcal M}$ restricted to the cotangent space of $\mathcal M$ is just the adjoint $(T\pi_{\mathcal M})^*: \mathcal T^* \mathcal M \to \mathcal T^*(\mathcal M \times \mathcal N)$ of the tangent map, i.e. $\pi_{\mathcal M}^*|_{\mathcal T^* \mathcal M} = (T\pi_{\mathcal M})^*.$ Thus $\pi_{\mathcal M}^* (\mathcal T^* \mathcal M) = (T\pi_{\mathcal M})^* (\mathcal T^* \mathcal M) \subset \mathcal T^*(\mathcal M \times N)$.
Since $\pi_{\mathcal M}$ is surjective, $T\pi_{\mathcal M}$ is also surjective and hence $(T\pi_{\mathcal M})^*$ is injective. Due to the finite dimensions of the involved spaces one concludes $\mathrm{dim}(\pi_{\mathcal M}^*(\mathcal T^* \mathcal M)) = \mathrm{dim}(\mathcal T^* \mathcal M) = \mathrm{dim}(\mathcal T \mathcal M) = \mathrm{dim}(\mathcal M).$
Repeating the same steps for $\mathcal N$ one obtains $\pi_{\mathcal N}^* (\mathcal T^* \mathcal N) = (T\pi_{\mathcal N})^* (\mathcal T^* \mathcal N) \subset \mathcal T^*(\mathcal M \times N)$ and $ \mathrm{dim}(\pi_{\mathcal N}^*(\mathcal T^* \mathcal N)) = \mathrm{dim}(\mathcal N).$
Finally$$ \begin{align} \mathrm{dim}(\pi_{\mathcal M}^*(\mathcal T^* \mathcal M)) + \mathrm{dim}(\pi_{\mathcal N}^*(\mathcal T^* \mathcal N)) &= \mathrm{dim}(\mathcal M) + \mathrm{dim}(\mathcal N)\\ &= \mathrm{dim}(\mathcal M \times \mathcal N)\\ &= \mathrm{dim}(\mathcal T^*(\mathcal M \times \mathcal N)), \end{align}$$where again the finite dimensionality of the involved spaces was made use of.
Let $\varphi \in \pi_{\mathcal M}^* (\mathcal T^* \mathcal M) \cap \pi_{\mathcal N}^* (\mathcal T^* \mathcal N)$ and let $u \in \mathcal T(\mathcal M \times \mathcal N)$, s.t. $\alpha$ and $u$ share the same base point in $\mathcal M \times \mathcal N$. By construction one has $\varphi = \pi_{\mathcal M}^* \alpha = \pi_{\mathcal N}^* \beta$ for some $\alpha \in \mathcal T^* \mathcal M,\,\beta \in \mathcal T^* \mathcal N.$ Moreover by Lemma 3 $u = v + w,\, v \in \mathrm{ker}(T\pi_{\mathcal M}),\, w \in \mathrm{ker}(T\pi_{\mathcal N})$.
Hence $$ \begin{align} \varphi(u) &= \varphi(v + w) = \varphi(v) + \varphi(w)\\ &= \pi_{\mathcal M}^* \alpha(v) + \pi_{\mathcal N}^* \beta(w) = \alpha(T\pi_{\mathcal M}v) + \beta(T\pi_{\mathcal N}w)\\ &= \alpha(0) + \beta(0) = 0 \end{align}$$ and therefore $\varphi = 0$. Hence $\pi_{\mathcal M}^* (\mathcal T^* \mathcal M) \cap \pi_{\mathcal N}^* (\mathcal T^* \mathcal N) = \{0\}.$$\square$
Lemma 3. $\mathcal T(\mathcal M \times N) = \mathrm{ker}(T\pi_{\mathcal M}) \oplus \mathrm{ker}(T\pi_{\mathcal n}).$
Remark: To show this I use the 'a vector is an equivalence class of curves' interpretation. I'm not particularly happy with this part. Is there a better / more elegant way? Maybe using derivations instead?
Let $u \in \mathrm{ker}(T_{(p,q)}\pi_{\mathcal M}) \cap \mathrm{ker}(T_{(p,q)}\pi_{\mathcal N})$. Then $u = [\gamma]$ is an equivalence class of curves $\gamma: ]-\varepsilon, \varepsilon[ \to \mathcal M \times \mathcal N$ which satisfy $\gamma(0) = (p, q),\; \left.\frac{\mathrm d}{\mathrm d t}\right|_{t=0}\pi_{\mathcal M}\circ\gamma = 0$ and $\left.\frac{\mathrm d}{\mathrm d t}\right|_{t=0}\pi_{\mathcal N}\circ\gamma = 0.$ Buț then it must also contain the constant curve $c \equiv (q, p)$, which can only be the case if $u = [c] = 0.$ Hence $\mathrm{ker}(T_{(p,q)}\pi_{\mathcal M}) \cap \mathrm{ker}(T_{(p,q)}\pi_{\mathcal N}) = \{0\}.$
Finally since $T\pi_{\mathcal M}$ is surjective one has $T\pi_{\mathcal M}(\mathcal {T}(\mathcal M \times \mathcal N)) = \mathcal {TM}.$ Hence $\mathrm{dim}(\mathrm{ker}(T\pi_{\mathcal M})) = \mathrm{dim}(\mathcal M \times \mathcal N) - \mathrm{dim}(\mathcal M) = \mathrm{dim}(\mathcal N),$ and similarly $\mathrm{dim}(\mathrm{ker}(T\pi_{\mathcal N})) = \mathrm{dim}(\mathcal M)$ so that $\mathrm{dim}(\mathrm{ker}(T\pi_{\mathcal M})) + \mathrm{dim}(\mathrm{ker}(T\pi_{\mathcal M})) = \mathrm{dim}(\mathcal M \times \mathcal N).\square$
