# The Unapologetic Mathematician

## Identifying Tensor Fields

Just as for vector fields, we need a good condition to identify tensor fields in the wild. And the condition we will use is similar: if $T$ is a smooth tensor field of type $(r,s)$, then for any coordinate patch $(U,x)$ in the domain of $T$, we should be able to write out

$\displaystyle T\vert_U=\sum\limits_{i_1,\dots,i_r,j_1,\dots,j_s=1}^nT^{i_1\dots i_r}{}_{j_1\dots j_s}\frac{\partial}{\partial x^{i_1}}\otimes\dots\otimes\frac{\partial}{\partial x^{i_r}}\otimes dx^{j_1}\otimes\dots\otimes dx^{j_s}$

for some smooth functions $T^{i_1\dots i_r}{}_{j_1\dots j_s}$ on $U$. Conversely, this formula defines a smooth tensor field on $U$.

Indeed, we can find these coefficient functions by evaluation:

$\displaystyle T^{i_1\dots i_r}{}_{j_1\dots j_s}=T\left(dx^{i_1},\dots,dx^{i_r},\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_s}}\right)$

for using this definition, if we plug these coordinate vector fields and coordinate covector fields into either the left or the right side of the expression above we will get the same answer. Any vector or covector fields on $U$ can be written as a linear combination of these coordinate fields with smooth functions as coefficients, and the multilinear properties of tensors will ensure that both sides get the same value no matter what fields we evaluate them on.

Similarly, if $\alpha$ is a differential $k$-form and $(U,x)$ is a coordinate patch within its domain, then we can write

$\displaystyle\alpha\vert_U=\sum\limits_{1\leq i_1<\dots

for some smooth functions $\alpha_{i_1\dots i_k}$ on $U$. The proof in this case is similar, following from the definition

$\displaystyle\alpha_{i_1\dots i_k}=\alpha\left(\frac{\partial}{\partial x^{i_1}},\dots,\frac{\partial}{\partial x^{i_k}}\right)$

In this case we can pick the indices to be strictly increasing because of the antisymmetry of the tensors.

July 7, 2011 - Posted by | Differential Topology, Topology

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