The Unapologetic Mathematician

Mathematics for the interested outsider

Tensor Fields and Multilinear Maps

A tensor field T over a manifold M gives us a tensor T_p at each point p\in M. And we know that T_p can be considered as a multilinear map. Specifically, if T is a tensor field of type (r,s), then we find

\displaystyle T_p\in\mathcal{T}_pM^{\otimes r}\otimes\mathcal{T}^*_pM^{\otimes s}

which we can interpret as a multilinear map:

\displaystyle T_p:\mathcal{T}_pM^{\times r}\times\mathcal{T}^*_pM^{\times s}\to\mathbb{R}

where multilinearity means that T_p is linear in each variable separately.

As we let p vary over M, we can interpret T as defining a function which takes r vector fields and s covector fields and gives a function. Explicitly:


And, in particular, this function is multilinear over \mathcal{O}M. That is,


And a similar calculation holds for any of the other variables, vector or covector.

So each tensor field gives us a multilinear function T:\mathfrak{X}M^{\times r}\times\mathfrak{X}^*M^{\times s}\to\mathcal{O}M, and this multilinearity is not only true over \mathbb{R} but over \mathcal{O}M as well.

Conversely, let T:\mathfrak{X}M^{\times r}\times\mathfrak{X}^*M^{\times s}\to\mathcal{O}M be an \mathbb{R}-multilinear function. If it’s also linear over \mathcal{O}M in each variable, then it “lives locally”. That is, if X_i(p)=Y_i(p) and \alpha^j(p)=\beta^j(p) then


and so at each p there is some tensor T_p\in T^r_s\left(\mathcal{T}_pM\right) so that T is a tensor field.

This is as distinguished from things like differential operators — X\to L_Y(X)^1, for instance — which fail both sides. On the one side, we can calculate


which picks up an extra term. It’s \mathbb{R}-linear but not \mathcal{O}M-linear. On the other side, the value of this function at p doesn’t just depend on the value of X at p, but on how X changes through p. That is, this operator does not “live locally”, and is not a tensor field.

To prove this assertion, it will suffice to deal with the case where T takes a single vector variable X, and we only need to verify that if X_p=0 then \left[T(X)\right](p)=0. Let (U,x) be a chart around p, and write

\displaystyle X=\sum\limits_{i=1}^nf^i\frac{\partial}{\partial x^i}

where by assumption each f^i(p)=0. We let V be a neighborhood of p whose closure is contained in U. We know we can find a smooth bump function \phi supported in U and with \phi(q)=1 on \bar{V}.

Now we define vector fields X_i=\phi\frac{\partial}{\partial x^i} on U and 0 on M\setminus U. Similarly we define g^i=\phi f^i on U and 0 on M\setminus U. Then we can write

\displaystyle X=\phi^2X+(1-\phi^2)X=\sum\limits_{i=1}^ng^iX_i+(1-\phi^2)X

and thus


as asserted.


July 9, 2011 - Posted by | Differential Topology, Topology

1 Comment »

  1. […] property onto the pile. Notice, though, how this condition is different from the way we said that tensor fields live locally. In this case we need to know that vanishes in a whole neighborhood, not just at […]

    Pingback by The Uniqueness of the Exterior Derivative « The Unapologetic Mathematician | July 19, 2011 | Reply

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