"Smooth" tensor field

Question

I am learning differential geometry and I have a question on a definition.

We call a tensor field a function that to a point on the manifold ($p\in M$), we associate smoothly a tensor.

But what does smoothly mean here?

In the book we defined a smooth function on a manifold $M$ to another $N$ as a function $f : M \rightarrow N $ which verifies : $ \phi \circ f \circ \psi^{-1} $ is a $C^{\infty}$ function from $\mathbb{R}^m$ to $\mathbb{R}^n$ (where $m$ and $n$ are the dimensions of the manifold $M$ and $N$ and $\phi$ and $\psi$ maps functions on $M$ and $N$.

But a tensor is not a function $f : M \rightarrow N $.

Answer 1

Let $\phi: U \subset M \to \mathbb R^n$ be a coordinate chart, and let $(x_1, \dots, x_n)$ be the corresponding coordinate functions. Write down the tensor field $T$ with respect to these coordinates: $$ T = T(x)^{\mu_1 \dots \mu_p}_{\nu_1 \dots \nu_q} \frac{\partial}{\partial x^{\mu_1}} \otimes \dots \otimes \frac{\partial}{\partial x^{\mu_p}} \otimes dx^{\nu_1} \otimes \dots \otimes dx^{\nu_q}.$$ The tensor field $T$ is smooth on the open set $U$ if and only if each $T(x)^{\mu_1 \dots \mu_p}_{\nu_1 \dots \nu_q}$ is a smooth function from $\phi(U)\subset \mathbb R^n$ to $\mathbb R$. To check that $T$ is smooth globally, you should find an atlas for $M$, and verify this condition on every chart in the atlas.

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Here is an alternative, equivalent definition: If $V$ is a vector field, then $V$ is smooth if and only if, for every smooth $f : M \to \mathbb R$, the function $V(f) : M \to \mathbb R$ is also smooth. (To see that this implies my definition, just apply this with $f$ equal to a coordinate function for a chart, multiplied by a bump function in order to extend it by zero outside of the chart. The converse holds by the fact that $V(f) (x)= V^\mu (x)\frac{\partial f}{\partial x^{\mu}}(x)$ in local coordinates.) If $\omega$ is a covector field, then $\omega$ is smooth if and only if, for every smooth vector field $X$, the function $\omega(X): M \to \mathbb R$ is smooth. (To see that this implies my definition, apply this with $X$ equal to the vector field $\partial/\partial x^{\mu_1}$ times a bump function. The converse holds because $\omega(X) (x)= \omega_\mu (x)X^\mu(x)$ in local coordinates.) It is easy to generalise this to arbitrary tensor fields.

Answer 2

$\newcommand{\dd}{\partial}$By hypothesis, $M$ is a smooth manifold. The definition of smoothness for a tensor (field) $T$ means: Fix an arbitrary coordinate system $(x^{i})$, in a neighborhood of $p$, belonging to the smooth atlas of $M$, and express $T$ in terms of the coordinate fields $\dd_{i} = \dd/\dd x_{i}$ and coordinate differentials $dx^{i}$. In multi-index notation: $$ T = \sum T_{I}^{J}\, dx^{I}\, \frac{\dd}{\dd x^{J}}. $$ To say $T$ is smooth is to say the component functions $T_{I}^{J}$ are smooth functions. (Naturally, one has to verify that if the components are smooth with respect to one coordinate system, they're smooth with respect to an arbitrary coordinate system. This is an easy consequence of the transformation rules for tensor components and the fact that change of coordinates maps are smooth. Note also that this definition reduces "as expected" if $M$ is an open subset of $\mathbf{R}^{n}$.)

Answer 3

Every bundle is a manifold and a tensor field is indeed a mapping between manifolds. Let ${M}$ be a smooth manifold and ${E}{M}$ be a tensor bundle over ${M}$ with the bundle projection $\pi:{E}{M}\longrightarrow{M}$. One then denotes the set of all smooth tensor fields by $\Gamma^{\infty}({M},{E}{M})$ which is defined as \begin{align*} \Gamma^{\infty}({M},{E}{M})=\lbrace{\,}{T}\in{C}^{\infty}({M},{E}{M}){\,}{\,}\mathrm{such}{\,}\mathrm{that}{\,}{\,}\pi\circ{T}=\mathrm{id}{\,}\rbrace{\,}{,} \end{align*} where the composition of the bundle projection after the field ensures that each point in the base manifold is mapped to the corresponding fiber within the bundle.

Answer 4

The other answers are excellent, but here is a less formal explanation:

A scalar field is a function that maps each point p in M to a value f(p) in R. It is intuitively clear what "smooth" means for a scalar field - it means f(q) is "close" to f(p) in R whenever q is "close" to p in M.

A vector field is a function that maps each point p in M to a vector f(p) in the tangent space to M at P, which we denote by $T_p$. Again, it is intuitively clear what "smooth" means for a vector field - it means f(q) is "close" to f(p) whenever q is "close" to p. The only complication is that we need some way of comparing vectors in the tangent space $T_q$ with vectors in the tangent space $T_p$. This comparison is given to us for free if M is embedded in an ambient space $R^m$, but it can be defined in more general cases too.

A tensor field just extends these notions to functions from M to tensors that live in a product of tangent and co-tangent spaces - the exact product depending on the type of the tensor. As long as we have a way of comparing objects in tangent and co-tangent spaces at nearby points p and q in M, then we can intuitively define the notion of a smooth tensor field.