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



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