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So, I know that tangent bundle of a product manifold $M \times N$ splits in a sum $$ T_{(x,y)}(M \times N) = T_xM \oplus T_yN, $$ so that is obvious that the sum $X \oplus Y$ of smooth vector fields $X \in \mathcal{T}(M)$ and $Y \in \mathcal{T}(N)$ is a smooth vector field of $M \times N$. I've been told that, though not every vector field in $\mathcal{T}(M \times N)$ is a sum, locally one can always find one such decomposition, which will in turn be unique due the fact the sum is a direct one.

How can I show that this decomposition exists locally? More than that, if $X = X_1 + X_2$ is the decomposition, is there a way the express the coordinate functions of $X_1$ and $X_2$ in terms of those of $X$?

First I thought about taking two frames that locally spam $TM$ and $TN$ and write down $X$ using them, but then the coordinate functions are of the form $X^i: M \times N \to \mathbb R$, and the vector field components in each subspace are not exactly fields of $M$ and $N$ because their coordinate functions don't have the right domains. Is there another better way to see this decomposition holds locally?

deabo
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1 Answers1

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This is totally false. Indeed, a vector field which is locally of the form $X\oplus Y$ is also globally of that form (the local $X$'s and $Y$'s will always glue together, since they are unique if they exist). Not every vector field on $M\times N$ has this form, since the $TM$ component of a vector field may change between points with the same $M$ coordinate.

For a really simple explicit example, let $M=N=\mathbb{R}$ and identify vector fields on $M$ and $N$ with functions $\mathbb{R}\to\mathbb{R}$ and vector fields on $M\times N$ with functions $\mathbb{R}^2\to\mathbb{R}^2$. Then given two such functions $X,Y:\mathbb{R}\to\mathbb{R}$, their sum $X\oplus Y$ is identified with the function $F(s,t)=(X(s),Y(t))$. Obviously not every smooth function $\mathbb{R}^2\to\mathbb{R}^2$ has this form (e.g., the function $F(s,t)=(t,s)$ does not).

Note that the post you link to does not claim any such thing. Instead it claims a vector field can locally be written as a linear combination of vector fields of the form $X\oplus 0$ or $Y\oplus 0$ with coefficients that are smooth functions on $M\times N$. Those coefficients are crucial, since they can be smooth functions which genuinely live on the product and don't come from either coordinate alone. Allowing such coefficients, the conclusion is trivial. Indeed, choosing local coordinates on $M\times N$ that are a product of local coordinates on $M$ and local coordinates on $N$, every vector field on $M\times N$ is locally a linear combination of the coordinate vector fields (with smooth functions as coefficients). The coordinate vector fields each have the form $X\oplus 0$ or $0\oplus Y$ (the coordinate vector fields for coordinates which come from $M$ are just $X\oplus 0$ where $X$ is the corresponding coordinate vector field on $M$, and similarly for the coordinates that come from $N$).

Eric Wofsey
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  • Ok, there is some nuance here then. in the link's question we're not decomposing $X$ and then looking at the Riemannian metric at each coordinate, but rather writing $X$ as a sum of components and then taking the inner product of each component as if it was a vector field on $M$ or $N$, because pointwise the decomposition always hold, is that right? – deabo Apr 11 '19 at 12:56
  • Your last statement confused me a little. When you say "the coordinate vector fields for coordinates which come from M are just $X \oplus 0$ where $X$ is the corresponding coordinate vector field on $M$, and similarly for the coordinates that come from $N$" don't mean exactly that we are decomposing $X$ as a direct sum of vector fields in $M$ and $N$? – deabo Apr 11 '19 at 12:58
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    The link's question is asking about the Riemannian connection, not the metric. In my last statement, $X$ is by definition a vector field on $M$, so I don't know what you're talking about. – Eric Wofsey Apr 11 '19 at 13:58
  • The conection appearing in the exercise comes from a riemannian metric $g(X,Y) = g_1(X_1,Y_1) + g_2(X_2, Y_2)$, where $X_i$ and $Y_i$ are the components as you described and $g_i$ are the metrics of $M$ and $N$. What I must do is to show this is indeed a metric and that its riemannian connection is as in the link, but I was stucked with the decomposition thing. – deabo Apr 11 '19 at 15:34
  • But when I write $X = (X_1 \oplus 0) \oplus (0 \oplus X_2)$ as you did, and each $X_i$ corresponds to a vector field on $M$ or $N$, why is isn't this a decomposition of $X$ as a direct sum of fields on $M$ and $N$? We may not explicitly know these fields' oordinates, but they are well defined fields on their respective manifolds, right? – deabo Apr 11 '19 at 15:40
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    I never wrote anything like $X = (X_1 \oplus 0) \oplus (0 \oplus X_2)$. I assumed you were using $\oplus$ for the external direct sum of vector fields, so when $X$ and $Y$ are vector fields on $M$ and $N$, $X\oplus Y$ is the vector field on $M\times N$ which at a point $(p,q)$ is given by $(X(p),Y(q))$ when you identify $T_{(p,q)}(M\times N)$ with $T_pM\times T_qN$. – Eric Wofsey Apr 11 '19 at 15:47
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    Note that the metric and the connection are very different beasts. For the metric, you only care about the values of your vector fields at a single point at a time. When looking at just one point $(p,q)\in M\times N$, every tangent vector can be written as the sum of a tangent vector from $M$ and a tangent vector from $N$ using the natural isomorphism $T_{(p,q)}(M\times N)\cong T_pM\times T_pN$. On the other hand, to talk about the connection you really need vector fields, not just tangent vectors at a single point. – Eric Wofsey Apr 11 '19 at 15:54