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The vector space $U_x$ will be infinite dimensional, so it's not immediately clear what $\Gamma^\infty(M,U)$ denotes. I assume you mean $\Gamma^\infty(M,U):=\bigoplus_k \Gamma^\infty(M,U^k)$ where $U^k_x:=\mathcal{S}^k/I_x\mathcal{S}^k$. Then the question might be: Why is $\mathcal{S}^k$ isomorphic to $\Gamma^\infty(M,U^k)$? When $\mathcal{S}^k$ is projective an finitely generated over $C^\infty(M)$ (which I think it is in your setting) the answer is the Serre-Swan theorem. You can find a proof for the $C^\infty$ setting in the book Nestruev, Smooth manifolds and observables.

Added in response to the comment:

You wrote:

Serre-Swan's theorem gives you that $\mathcal{S}^k$ can be identified with $\Gamma^\infty(M,V)$ for some vector bundle $V$ (provided we can show that $\mathcal{S}^k$ is finitely generated and projective). On the other hand, here we would like to have a concrete description of $V$ as $U$ where the fiber over $x$ is $\mathcal{S}^k/I_x\mathcal{S}^k$.

That concrete description always follows from an identification $\Gamma^\infty(M,V)=\mathcal{S}^k$. To be precise: from a vector bundle $V$, a $C^\infty(M)$-module $\mathcal{P}$ and a $C^\infty(M)$-module isomorphism $\phi: \Gamma^\infty(M,V)\to \mathcal{P}$, we always obtain a natural identification $V_x= \mathcal{P}/I_x\mathcal{P}$. This identification is constructed as follows: given $x\in M$, let's consider $\mathbb{R}$ as a $C^\infty(M)$-module via the canonical identification $C^\infty(M)/I_x = \mathbb{R}$. Next we have a natural identification $\mathcal{P}/I_x\mathcal{P}=\mathcal{P}\otimes_{C^\infty M} \mathbb{R}$, given in the direction $ \mathcal{P}/I_x\mathcal{P}\to \mathcal{P}\otimes_{C^\infty M} \mathbb{R} $ by $$ p \text{ mod } I_x\mathcal{P}\mapsto p\otimes 1 $$ and in the other direction by $$ p\otimes c \mapsto p\cdot c \text{ mod } I_x\mathcal{P}. $$ (I skip the details of verifying that these maps are well defined and yield the identity when composed. Feel free to ask if something is not clear.)

Hence, tensoring the isomorphism $\phi$ with $\mathbb{R}$ we obtain an identification $$ \alpha: \Gamma^\infty(M,V)/I_x\Gamma^\infty(M,V) \to \mathcal{P} /I_x\mathcal{P}. $$ As last step, it's not hard to see that $\Gamma^\infty(M,V)/I_x\Gamma^\infty(M,V)$ is naturally identified with $V_x$ for any vector bundle $V$ (Corollary 11.9 in Nestruev).

To see that your concrete module $\mathcal{S}^k$ is finitely generated and projective we can construct an isomorphism $\mathcal{S}^k = \mathcal{Q}^*\otimes_{C^\infty(M)} \mathcal{Q}\otimes_{C^\infty(M)} S^k D$, where I'm using the following abbreviations

• $\mathcal{Q} =\Gamma^\infty(M,E)$
• $D$ denotes the module of vector fields on $M$, i.e. sections of the tangent bundle,
• $S^k$ denotes the $k$-th symmetric product of a $C^\infty(M)$-module.
• $\mathcal{Q}^*=\text{Hom}_{C^\infty M}(Q,C^\infty M)$,

In one direction this iso is defined by \begin{align} \mathcal{Q}^*\otimes_{C^\infty M} \mathcal{Q}\otimes_{C^\infty M} S^k D &\to \mathcal{P}^k/\mathcal{P}^{k-1}\\ r\otimes s \otimes (X_1\cdots X_k) &\mapsto s\circ X_1\circ\ldots\circ X_k \circ r \quad \text{mod} \quad \mathcal{P}^{k-1} \end{align}

To make sense of the composition with $s$, use the canonical identification of a module $\mathcal{Q}$ with $\text{Hom}_{C^\infty M}(C^\infty M, \mathcal{Q})$.

The other direction is also not to hard to define, but slightly more involved: start with the observation that given $(\nabla \text{ mod } \mathcal{P}^{k-1}) \in \mathcal{P}^k/\mathcal{P}^{k-1}$ and $a_1,\ldots,a_k \in C^\infty M$, then $[\cdots [P,a_1],a_2]\cdots ,a_k]\in \mathcal{Q}^*\otimes \mathcal{Q}$ is well-defined (independent of the representative $\nabla\in \mathcal{P}^k$) and multilinear symmetric and a derivation in the $a_i$'s. From there you can use the universal derivation $d: A \to \Lambda^1$ (differential one forms) to show that you get a well defined map from $ \mathcal{P}^k/\mathcal{P}^{k-1}$ to $\mathcal{Q}^*\otimes_{C^\infty(M)} \mathcal{Q}\otimes_{C^\infty(M)} S^k D$.

This identification not only shows that $\mathcal{S}^k$ is projective and finitely generated, it gives a more concrete description of fibers $\mathcal{S}^k/I_x\mathcal{S}^k$: they are symbols $S^k T_xM\otimes_\mathbb{R} \text{Hom}_\mathbb{R}(E_x,E_x)$, as mentioned in Peter Michor's answer.

The vector space $U_x$ will be infinite dimensional, so it's not immediately clear what $\Gamma^\infty(M,U)$ denotes. I assume you mean $\Gamma^\infty(M,U):=\bigoplus_k \Gamma^\infty(M,U^k)$ where $U^k_x:=\mathcal{S}^k/I_x\mathcal{S}^k$. Then the question might be: Why is $\mathcal{S}^k$ isomorphic to $\Gamma^\infty(M,U^k)$? When $\mathcal{S}^k$ is projective an finitely generated over $C^\infty(M)$ (which I think it is in your setting) the answer is the Serre-Swan theorem. You can find a proof for the $C^\infty$ setting in the book Nestruev, Smooth manifolds and observables.

Added in response to the comment:

You wrote:

Serre-Swan's theorem gives you that $\mathcal{S}^k$ can be identified with $\Gamma^\infty(M,V)$ for some vector bundle $V$ (provided we can show that $\mathcal{S}^k$ is finitely generated and projective). On the other hand, here we would like to have a concrete description of $V$ as $U$ where the fiber over $x$ is $\mathcal{S}^k/I_x\mathcal{S}^k$.

That concrete description always follows from an identification $\Gamma^\infty(M,V)=\mathcal{S}^k$. To be precise: from a vector bundle $V$, a $C^\infty(M)$-module $\mathcal{P}$ and a $C^\infty(M)$-module isomorphism $\phi: \Gamma^\infty(M,V)\to \mathcal{P}$, we always obtain a natural identification $V_x= \mathcal{P}/I_x\mathcal{P}$. This identification is constructed as follows: given $x\in M$, let's consider $\mathbb{R}$ as a $C^\infty(M)$-module via the canonical identification $C^\infty(M)/I_x = \mathbb{R}$. Next we have a natural identification $\mathcal{P}/I_x\mathcal{P}=\mathcal{P}\otimes_{C^\infty M} \mathbb{R}$, given in the direction $ \mathcal{P}/I_x\mathcal{P}\to \mathcal{P}\otimes_{C^\infty M} \mathbb{R} $ by $$ p \text{ mod } I_x\mathcal{P}\mapsto p\otimes 1 $$ and in the other direction by $$ p\otimes c \mapsto p\cdot c \text{ mod } I_x\mathcal{P}. $$ (I skip the details of verifying that these maps are well defined and yield the identity when composed. Feel free to ask if something is not clear.)

Hence, tensoring the isomorphism $\phi$ with $\mathbb{R}$ we obtain an identification $$ \alpha: \Gamma^\infty(M,V)/I_x\Gamma^\infty(M,V) \to \mathcal{P} /I_x\mathcal{P}. $$ As last step, it's not hard to see that $\Gamma^\infty(M,V)/I_x\Gamma^\infty(M,V)$ is naturally identified with $V_x$ for any vector bundle $V$ (Corollary 11.9 in Nestruev).

To see that your concrete module $\mathcal{S}^k$ is finitely generated and projective we can construct an isomorphism $\mathcal{S}^k = \mathcal{Q}^*\otimes_{C^\infty(M)} \mathcal{Q}\otimes_{C^\infty(M)} S^k D$, where I'm using the following abbreviations

• $\mathcal{Q} =\Gamma^\infty(M,E)$
• $D$ denotes the module of vector fields on $M$, i.e. sections of the tangent bundle,
• $S^k$ denotes the $k$-th symmetric product of a $C^\infty(M)$-module.
• $\mathcal{Q}^*=\text{Hom}_{C^\infty M}(Q,C^\infty M)$,

In one direction this iso is defined by \begin{align} \mathcal{Q}^*\otimes_{C^\infty M} \mathcal{Q}\otimes_{C^\infty M} S^k D &\to \mathcal{P}^k/\mathcal{P}^{k-1}\\ r\otimes s \otimes (X_1\cdots X_k) &\mapsto s\circ X_1\circ\ldots\circ X_k \circ r \quad \text{mod} \quad \mathcal{P}^{k-1} \end{align}

To make sense of the composition with $s$, use the canonical identification of a module $\mathcal{Q}$ with $\text{Hom}_{C^\infty M}(C^\infty M, \mathcal{Q})$.

The other direction is also not to hard to define, but slightly more involved: start with the observation that given $(\nabla \text{ mod } \mathcal{P}^{k-1}) \in \mathcal{P}^k/\mathcal{P}^{k-1}$ and $a_1,\ldots,a_k \in C^\infty M$, then $[\cdots [\nabla,a_1],a_2]\cdots ,a_k]\in \mathcal{Q}^*\otimes \mathcal{Q}$ is well-defined (independent of the representative $\nabla\in \mathcal{P}^k$) and multilinear symmetric and a derivation in the $a_i$'s. From there you can use the universal derivation $d: A \to \Lambda^1$ (differential one forms) to show that you get a well defined map from $ \mathcal{P}^k/\mathcal{P}^{k-1}$ to $\mathcal{Q}^*\otimes_{C^\infty(M)} \mathcal{Q}\otimes_{C^\infty(M)} S^k D$.

This identification not only shows that $\mathcal{S}^k$ is projective and finitely generated, it gives a more concrete description of fibers $\mathcal{S}^k/I_x\mathcal{S}^k$: they are symbols $S^k T_xM\otimes_\mathbb{R} \text{Hom}_\mathbb{R}(E_x,E_x)$, as mentioned in Peter Michor's answer.

The vector space $U_x$ will be infinite dimensional, so it's not immediately clear what $\Gamma^\infty(M,U)$ denotes. I assume you mean $\Gamma^\infty(M,U):=\bigoplus_k \Gamma^\infty(M,U^k)$ where $U^k_x:=\mathcal{S}^k/I_x\mathcal{S}^k$. Then the question might be: Why is $\mathcal{S}^k$ isomorphic to $\Gamma^\infty(M,U^k)$? When $\mathcal{S}^k$ is projective an finitely generated over $C^\infty(M)$ (which I think it is in your setting) the answer is the Serre-Swan theorem. You can find a proof for the $C^\infty$ setting in the book Nestruev, Smooth manifolds and observables.

Added in response to the comment:

You wrote:

Serre-Swan's theorem gives you that $\mathcal{S}^k$ can be identified with $\Gamma^\infty(M,V)$ for some vector bundle $V$ (provided we can show that $\mathcal{S}^k$ is finitely generated and projective). On the other hand, here we would like to have a concrete description of $V$ as $U$ where the fiber over $x$ is $\mathcal{S}^k/I_x\mathcal{S}^k$.

That concrete description always follows from an identification $\Gamma^\infty(M,V)=\mathcal{S}^k$. To be precise: from a vector bundle $V$, a $C^\infty(M)$-module $\mathcal{P}$ and a $C^\infty(M)$-module isomorphism $\phi: \Gamma^\infty(M,V)\to \mathcal{P}$, we always obtain a natural identification $V_x= \mathcal{P}/I_x\mathcal{P}$. This identification is constructed as follows: given $x\in M$, let's consider $\mathbb{R}$ as a $C^\infty(M)$-module via the canonical identification $C^\infty(M)/I_x = \mathbb{R}$. Next we have a natural identification $\mathcal{P}/I_x\mathcal{P}=\mathcal{P}\otimes_{C^\infty M} \mathbb{R}$, given in the direction $ \mathcal{P}/I_x\mathcal{P}\to \mathcal{P}\otimes_{C^\infty M} \mathbb{R} $ by $$ p \text{ mod } I_x\mathcal{P}\mapsto p\otimes 1 $$ and in the other direction by $$ p\otimes c \mapsto p\cdot c \text{ mod } I_x\mathcal{P}. $$ (I skip the details of verifying that these maps are well defined and yield the identity when composed. Feel free to ask if something is not clear.)

Hence, tensoring the isomorphism $\phi$ with $\mathbb{R}$ we obtain an identification $$ \alpha: \Gamma^\infty(M,V)/I_x\Gamma^\infty(M,V) \to \mathcal{P} /I_x\mathcal{P}. $$ As last step, it's not hard to see that $\Gamma^\infty(M,V)/I_x\Gamma^\infty(M,V)$ is naturally identified with $V_x$ for any vector bundle $V$ (Corollary 11.9 in Nestruev).

To see that your concrete module $\mathcal{S}^k$ is finitely generated and projective we can construct an isomorphism $\mathcal{S}^k = \mathcal{Q}^*\otimes_{C^\infty(M)} \mathcal{Q}\otimes_{C^\infty(M)} S^k D$, where I'm using the following abbreviations

• $\mathcal{Q} =\Gamma^\infty(M,E)$
• $D$ denotes the module of vector fields on $M$, i.e. sections of the tangent bundle,
• $S^k$ denotes the $k$-th symmetric product of a $C^\infty(M)$-module.
• $\mathcal{Q}^*=\text{Hom}_{C^\infty M}(Q,C^\infty M)$,

In one direction this iso is defined by \begin{align} \mathcal{Q}^*\otimes_{C^\infty M} \mathcal{Q}\otimes_{C^\infty M} S^k D &\to \mathcal{P}^k/\mathcal{P}^{k-1}\\ r\otimes s \otimes (X_1\cdots X_k) &\mapsto s\circ X_1\circ\ldots\circ X_k \circ r \quad \text{mod} \quad \mathcal{P}^{k-1} \end{align}

The other direction is also not to hard to define, but slightly more involved.

This identification not only shows that $\mathcal{S}^k$ is projective and finitely generated, it gives a more concrete description of fibers $\mathcal{S}^k/I_x\mathcal{S}^k$: they are symbols $S^k T_xM\otimes_\mathbb{R} \text{Hom}_\mathbb{R}(E_x,E_x)$, as mentioned in Peter Michor's answer.

The vector space $U_x$ will be infinite dimensional, so it's not immediately clear what $\Gamma^\infty(M,U)$ denotes. I assume you mean $\Gamma^\infty(M,U):=\bigoplus_k \Gamma^\infty(M,U^k)$ where $U^k_x:=\mathcal{S}^k/I_x\mathcal{S}^k$. Then the question might be: Why is $\mathcal{S}^k$ isomorphic to $\Gamma^\infty(M,U^k)$? When $\mathcal{S}^k$ is projective an finitely generated over $C^\infty(M)$ (which I think it is in your setting) the answer is the Serre-Swan theorem. You can find a proof for the $C^\infty$ setting in the book Nestruev, Smooth manifolds and observables.

Added in response to the comment:

You wrote:

Serre-Swan's theorem gives you that $\mathcal{S}^k$ can be identified with $\Gamma^\infty(M,V)$ for some vector bundle $V$ (provided we can show that $\mathcal{S}^k$ is finitely generated and projective). On the other hand, here we would like to have a concrete description of $V$ as $U$ where the fiber over $x$ is $\mathcal{S}^k/I_x\mathcal{S}^k$.

That concrete description always follows from an identification $\Gamma^\infty(M,V)=\mathcal{S}^k$. To be precise: from a vector bundle $V$, a $C^\infty(M)$-module $\mathcal{P}$ and a $C^\infty(M)$-module isomorphism $\phi: \Gamma^\infty(M,V)\to \mathcal{P}$, we always obtain a natural identification $V_x= \mathcal{P}/I_x\mathcal{P}$. This identification is constructed as follows: given $x\in M$, let's consider $\mathbb{R}$ as a $C^\infty(M)$-module via the canonical identification $C^\infty(M)/I_x = \mathbb{R}$. Next we have a natural identification $\mathcal{P}/I_x\mathcal{P}=\mathcal{P}\otimes_{C^\infty M} \mathbb{R}$, given in the direction $ \mathcal{P}/I_x\mathcal{P}\to \mathcal{P}\otimes_{C^\infty M} \mathbb{R} $ by $$ p \text{ mod } I_x\mathcal{P}\mapsto p\otimes 1 $$ and in the other direction by $$ p\otimes c \mapsto p\cdot c \text{ mod } I_x\mathcal{P}. $$ (I skip the details of verifying that these maps are well defined and yield the identity when composed. Feel free to ask if something is not clear.)

Hence, tensoring the isomorphism $\phi$ with $\mathbb{R}$ we obtain an identification $$ \alpha: \Gamma^\infty(M,V)/I_x\Gamma^\infty(M,V) \to \mathcal{P} /I_x\mathcal{P}. $$ As last step, it's not hard to see that $\Gamma^\infty(M,V)/I_x\Gamma^\infty(M,V)$ is naturally identified with $V_x$ for any vector bundle $V$ (Corollary 11.9 in Nestruev).

To see that your concrete module $\mathcal{S}^k$ is finitely generated and projective we can construct an isomorphism $\mathcal{S}^k = \mathcal{Q}^*\otimes_{C^\infty(M)} \mathcal{Q}\otimes_{C^\infty(M)} S^k D$, where I'm using the following abbreviations

• $\mathcal{Q} =\Gamma^\infty(M,E)$
• $D$ denotes the module of vector fields on $M$, i.e. sections of the tangent bundle,
• $S^k$ denotes the $k$-th symmetric product of a $C^\infty(M)$-module.
• $\mathcal{Q}^*=\text{Hom}_{C^\infty M}(Q,C^\infty M)$,

In one direction this iso is defined by \begin{align} \mathcal{Q}^*\otimes_{C^\infty M} \mathcal{Q}\otimes_{C^\infty M} S^k D &\to \mathcal{P}^k/\mathcal{P}^{k-1}\\ r\otimes s \otimes (X_1\cdots X_k) &\mapsto s\circ X_1\circ\ldots\circ X_k \circ r \quad \text{mod} \quad \mathcal{P}^{k-1} \end{align}

To make sense of the composition with $s$, use the canonical identification of a module $\mathcal{Q}$ with $\text{Hom}_{C^\infty M}(C^\infty M, \mathcal{Q})$.

The other direction is also not to hard to define, but slightly more involved: start with the observation that given $(\nabla \text{ mod } \mathcal{P}^{k-1}) \in \mathcal{P}^k/\mathcal{P}^{k-1}$ and $a_1,\ldots,a_k \in C^\infty M$, then $[\cdots [P,a_1],a_2]\cdots ,a_k]\in \mathcal{Q}^*\otimes \mathcal{Q}$ is well-defined (independent of the representative $\nabla\in \mathcal{P}^k$) and multilinear symmetric and a derivation in the $a_i$'s. From there you can use the universal derivation $d: A \to \Lambda^1$ (differential one forms) to show that you get a well defined map from $ \mathcal{P}^k/\mathcal{P}^{k-1}$ to $\mathcal{Q}^*\otimes_{C^\infty(M)} \mathcal{Q}\otimes_{C^\infty(M)} S^k D$.

This identification not only shows that $\mathcal{S}^k$ is projective and finitely generated, it gives a more concrete description of fibers $\mathcal{S}^k/I_x\mathcal{S}^k$: they are symbols $S^k T_xM\otimes_\mathbb{R} \text{Hom}_\mathbb{R}(E_x,E_x)$, as mentioned in Peter Michor's answer.

The vector space $U_x$ will be infinite dimensional, so it's not immediately clear what $\Gamma^\infty(M,U)$ denotes. I assume you mean $\Gamma^\infty(M,U):=\bigoplus_k \Gamma^\infty(M,U^k)$ where $U^k_x:=\mathcal{S}^k/I_x\mathcal{S}^k$. Then the question reduces to: Why is $\mathcal{S}^k$ isomorphic to $\Gamma^\infty(M,U^k)$? When $\mathcal{S}^k$ is projective an finitely generated over $C^\infty(M)$ (which I think it is in your setting) the answer is the Serre-Swan theorem. I think you can find a proof for the $C^\infty$ setting in the book Nestruev, Smooth manifolds and observables.

Added after the comments: Since you were not very specific about the setting in your question, I'm going to make the following assumptions: your $M$ is a real smooth manifold and $C^\infty(M)$ denotes the algebra of real valued smooth functions. I assume that $\Gamma^\infty (M,V)$ denotes the $C^\infty(M)$-module of smooth sections of a finite dimensional real vector bundle $\pi:V\to M$. You mention that the bundle $E$ is complex, but I'll assume, when you talk of differential operators between $\Gamma^\infty(M,E)$, you mean differential operators in the real sense, i.e. $P:\Gamma^\infty(M,E)\to \Gamma^\infty(M,E)$ is a differential operator of order $k$, iff $[f,P]$ is a differential operator of order $k-1$ for every $f\in C^\infty(M)$. I believe that what I'm about to say has straightforward modifications in case you meant $C^\infty(M)$ to be complex valued functions, or $M$ to be complex analytic.

In your comment you wrote:

Serre-Swan's theorem gives you that $\mathcal{S}^k$ can be identified with $\Gamma^\infty(M,V)$ for some vector bundle $V$ ... On the other hand, here we would like to have a concrete description of $V$ as $U$ where the fiber over $x$ is $\mathcal{S}^k/I_x\mathcal{S}^k$.

The proof of Serre-Swan given in for example Nestruev (Theorem 11.32) indeed seems to only give some vector bundle, without showing that the construction is functorial. It's probably possible to show that the construction in Nestruev is functorial, but one could also address the concern in your comment directly, by defining the functor from projective finitely generated $C^\infty(M)$-modules to vector bundles over $M$ as $$\mathcal{Q} \mapsto \left(\bigcup_{x\in M} \mathcal{Q}/I_x \mathcal{Q}\to M\right).$$ To make sense of this, we still need to explain what topology and smooth structure to put on the disjoint union $\bigcup_{x\in M} \mathcal{Q}/I_x \mathcal{Q}$. It should not be to hard to fill in the details with the other things you can find in the Nestruevs book. (Note that by saying what each fiber $U_x$ is in you're question, you had not defined a vector bundle $U$ yet.)

Finally, to show that $\mathcal{S}^k$ is finitely generated and projective you can construct an isomorphism $\mathcal{S}^k = \mathcal{Q}^*\otimes_{C^\infty(M)} \mathcal{Q}\otimes_{C^\infty(M)} S^k D$, where I'm using the following abbreviations

• $Q =\Gamma^\infty(M,E)$
• $D$ denotes the module of vector fields on $M$, i.e. sections of the tangent bundle.
• $S^k$ denotes the $k$-th symmetric product of a $C^\infty(M)$-module.
• $Q^*=\text{Hom}_{C^\infty(M)}(Q,C^\infty(M))$.

In one direction this iso is defined by \begin{align} \mathcal{Q}^*\otimes_{C^\infty(M)} \mathcal{Q}\otimes_{C^\infty(M)} S^k D &\to P^k/P^{k-1}\\ r\otimes s \otimes (X_1\cdots X_k) &\mapsto s\circ X_1\circ\ldots\circ X_k \circ r \quad \text{mod} \quad P^{k-1} \end{align}

The other direction of the iso is also not to hard to define, but is slightly more involved.

With this isomorphism we could have skipped all the talk of Serre-Swan, since it directly yields a description of the fiber $U^k_x$ of the bundle you are interested in. Namely it's canonically isomorphic to $S^k T_xM\otimes_\mathbb{R} \text{Hom}_\mathbb{R}(E_x,E_x)$, which are symbols, as Peter Michor mentions. (This final step requires the fact that pulling back modules (extension of scalars) commutes with tensor products.)

The vector space $U_x$ will be infinite dimensional, so it's not immediately clear what $\Gamma^\infty(M,U)$ denotes. I assume you mean $\Gamma^\infty(M,U):=\bigoplus_k \Gamma^\infty(M,U^k)$ where $U^k_x:=\mathcal{S}^k/I_x\mathcal{S}^k$. Then the question might be: Why is $\mathcal{S}^k$ isomorphic to $\Gamma^\infty(M,U^k)$? When $\mathcal{S}^k$ is projective an finitely generated over $C^\infty(M)$ (which I think it is in your setting) the answer is the Serre-Swan theorem. You can find a proof for the $C^\infty$ setting in the book Nestruev, Smooth manifolds and observables.

Added in response to the comment:

You wrote:

Serre-Swan's theorem gives you that $\mathcal{S}^k$ can be identified with $\Gamma^\infty(M,V)$ for some vector bundle $V$ (provided we can show that $\mathcal{S}^k$ is finitely generated and projective). On the other hand, here we would like to have a concrete description of $V$ as $U$ where the fiber over $x$ is $\mathcal{S}^k/I_x\mathcal{S}^k$.

That concrete description always follows from an identification $\Gamma^\infty(M,V)=\mathcal{S}^k$. To be precise: from a vector bundle $V$, a $C^\infty(M)$-module $\mathcal{P}$ and a $C^\infty(M)$-module isomorphism $\phi: \Gamma^\infty(M,V)\to \mathcal{P}$, we always obtain a natural identification $V_x= \mathcal{P}/I_x\mathcal{P}$. This identification is constructed as follows: given $x\in M$, let's consider $\mathbb{R}$ as a $C^\infty(M)$-module via the canonical identification $C^\infty(M)/I_x = \mathbb{R}$. Next we have a natural identification $\mathcal{P}/I_x\mathcal{P}=\mathcal{P}\otimes_{C^\infty M} \mathbb{R}$, given in the direction $ \mathcal{P}/I_x\mathcal{P}\to \mathcal{P}\otimes_{C^\infty M} \mathbb{R} $ by $$ p \text{ mod } I_x\mathcal{P}\mapsto p\otimes 1 $$ and in the other direction by $$ p\otimes c \mapsto p\cdot c \text{ mod } I_x\mathcal{P}. $$ (I skip the details of verifying that these maps are well defined and yield the identity when composed. Feel free to ask if something is not clear.)

Hence, tensoring the isomorphism $\phi$ with $\mathbb{R}$ we obtain an identification $$ \alpha: \Gamma^\infty(M,V)/I_x\Gamma^\infty(M,V) \to \mathcal{P} /I_x\mathcal{P}. $$ As last step, it's not hard to see that $\Gamma^\infty(M,V)/I_x\Gamma^\infty(M,V)$ is naturally identified with $V_x$ for any vector bundle $V$ (Corollary 11.9 in Nestruev).

To see that your concrete module $\mathcal{S}^k$ is finitely generated and projective we can construct an isomorphism $\mathcal{S}^k = \mathcal{Q}^*\otimes_{C^\infty(M)} \mathcal{Q}\otimes_{C^\infty(M)} S^k D$, where I'm using the following abbreviations

• $\mathcal{Q} =\Gamma^\infty(M,E)$
• $D$ denotes the module of vector fields on $M$, i.e. sections of the tangent bundle,
• $S^k$ denotes the $k$-th symmetric product of a $C^\infty(M)$-module.
• $\mathcal{Q}^*=\text{Hom}_{C^\infty M}(Q,C^\infty M)$,

In one direction this iso is defined by \begin{align} \mathcal{Q}^*\otimes_{C^\infty M} \mathcal{Q}\otimes_{C^\infty M} S^k D &\to \mathcal{P}^k/\mathcal{P}^{k-1}\\ r\otimes s \otimes (X_1\cdots X_k) &\mapsto s\circ X_1\circ\ldots\circ X_k \circ r \quad \text{mod} \quad \mathcal{P}^{k-1} \end{align}

The other direction is also not to hard to define, but slightly more involved.

This identification not only shows that $\mathcal{S}^k$ is projective and finitely generated, it gives a more concrete description of fibers $\mathcal{S}^k/I_x\mathcal{S}^k$: they are symbols $S^k T_xM\otimes_\mathbb{R} \text{Hom}_\mathbb{R}(E_x,E_x)$, as mentioned in Peter Michor's answer.