Let $M$ and $N$ be smooth manifolds of dimension $m$ and $n$ and for $p \in M$ denote
\begin{equation}
\mathcal{T}_p M = \lbrace \gamma \in C^{\infty}(I,M) | 0 \in I \land \gamma(0) = p \rbrace.
\end{equation}
Let $\sim$ be a equivalence relation on $\mathcal{T}_p M$ defined by
\begin{equation}
c_1 \sim c_2 : \iff \exists (U_{\varphi},\varphi) \in \mathcal{A}_M: p \in U_{\varphi} \land \frac{d}{dt}(\varphi \circ c_1)(0) = \frac{d}{dt}(\varphi \circ c_2)(0)
\end{equation}
where $\mathcal{A}_M$ is a atlas obtained by the smooth structure of $M$
then the tangent space $T_p M$ of $M$ at $p$ is defined by
\begin{equation}
T_p M := \mathcal{T}_p M / \sim.
\end{equation}
Now I am supposed to proove that:
\begin{equation}
T_{(p,q)}(M \times N) \cong T_pM \oplus T_qN
\end{equation}
My question now is rather simple, it isn about the acual exercise. I havent had a lot of contact with $\oplus$, what I heard back in the days was that we write $X = V \oplus W$ for 2 subspaces if $V \cap W = {0}$ and $X=V+W$. But in this case now, both tangent spaces are vectorspaces themselves and the addition is not defined since they are not subspaces of a commong vectorspace.
Reading on wikipedia explained me that the situation with $X = V \oplus W$ is an internal direct sum, and the situation I want to have is an external.
What does $T_pM \oplus T_qN$ mean? Can you please explain a bit about these direct sums?
Thank you, and sorry Orb for stealing your introduction, but you wrote it so nicely that I had to use it.
Best Answer
You can think of the direct sum $V \oplus W$ of two vector spaces $V$ and $W$ explicitly as the set of tuples $(a,b)$, where $a \in V$ and $b \in W$. Addition and scalar multiplication are defined component-wise.
The idea of a direct sum is to build a new vector space (or abelian group, or $R$-module, or whatever you want) from the ones you already have without introducing any relations between your vector spaces. If you have seen some category theory, the direct sum is the coproduct in the category of $k$-vector spaces. If you don't know what that means, feel free to ignore it.