What do tensor fields, both smooth and not smooth, correspond to?

Question

My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).

Proposition 21.11 says smooth tensor fields $\tilde T$, defined as smooth sections of $T^{(a,b)}M$, i.e. elements of $\Gamma(T^{(a,b)}M)$, bijectively correspond to $C^{\infty}M$-multi-linear maps $T: (\Omega^1M)^a \times \mathfrak X(M)^b \to C^{\infty}M$

Question: What do tensor fields $\tilde S$, smooth or not, defined as sections, smooth or not, of $T^{(a,b)}M$, i.e. elements of $A(T^{(a,b)}M) := \{\text{all sections of} \ T^{(a,b)}M \}$, bijectively correspond to?

My thoughts: The set $A(M):=\{\text{all maps} \ f : M \to \mathbb R\}$ can be made into a ring such that $C^{\infty}M$ is a subring of $A(M)$ i.e. $A(M)$ becomes a ring that is a superring of $C^{\infty}M$. Similarly, we have that $O^1M$ and $\mathcal X(M)$, the smooth-or-not supersets of, respectively, $\Omega^1M$ and $\mathfrak X(M)$, can be made into $\mathbb R$-superspaces and $C^{\infty}M$-supermodules, of, respectively, $\Omega^1M$ and $\mathfrak X(M)$.

Then $\tilde S$ bijectively corresponds to:

1. Guess 1: some $C^\infty M$-linear $S_1: (\Omega^1M)^a \times \mathfrak X(M)^b \to A(M)$

2. Guess 2: some $A(M)$-linear $S_2: (O^1M)^a \times \mathcal X(M)^b \to A(M)$

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Context:

1. I'm wondering what Proposition 22.7 would be simply for $\omega$ and (thus) $\nabla_X\omega$ not necessarily smooth.

2. But I think we can think of the above question in terms of 1-forms $\omega$:

$\omega \in \Omega^1M$ if and only if $\hat{\omega}: \mathfrak X(M) \to C^{\infty}M$ is $C^{\infty}M$-linear, where $\hat{\omega}(X) := \omega(X)$, where $\omega(X): M \to \mathbb R$, $\omega(X) \in C^{\infty}M$, $(\omega(X))(p) := \omega_p(X_p)$

Then, I think $\omega$, a 1-form on $M$, not necessarily smooth, bijectively corresponds to

• some $A(M)$-linear, $\hat{\omega}: \mathcal X(M) \to A(M)$

• or some $C^{\infty}M$-linear, $\hat{\omega}: \mathfrak X(M) \to A(M)$,

where, in either case, we still have $\hat{\omega}(X) := \omega(X)$ and $\omega(X): M \to \mathbb R$, not necessarily smooth, $(\omega(X))(p) := \omega_p(X_p)$

Answer 1

Your guess 2 is correct: rough tensor fields correspond to $A(M)$-multilinear maps (acting on rough vector fields/one-forms). Let's quickly prove it for one-forms.

Given a rough one-form $\omega \in O^1$, the corresponding map $\hat \omega:\mathcal X \to A$ defined by $(\hat \omega(V))_x = \omega(V_x)$ is $A$-linear since $$\hat \omega(fV)_x = \omega((fV)_x) = f(x)\omega (V_x)=(f\hat\omega(V))_x.$$

Conversely, if $\hat \omega : \mathcal X \to A$ is $A$-linear, then for any point $x$ the indicator function $1_x$ is in $A$; so by linearity we have $\hat \omega(1_xV) = 1_x \hat \omega(V).$ Evaluating this at $x$ we see that the value of $\hat \omega(V)$ at $x$ depends only on the value of $V$ at $x;$ so there is some $\omega_x : T_x \to \mathbb R$ such that $\hat \omega(V)_x = \omega(V_x)$ for all $V_x$. Gathering up these $\omega_x$ gives us the desired section $\omega \in O^1$.