Every noncompact manifold admits nonzero vector fields, or more generally, vector fields with any specified set of isolated zeros along with the behavior near that zero.

However, if you have information of the behavior of a vector field near infinity, or just in a neighborhood of the boundary of a compact set, there is an index theorem. Perhaps this is the case with your $\mathbb T^! \times \mathbb R$.

In the particular case of a cylinder, there is a simple way to calculate the index. Take any compact subcylinder delimited by two circles. Map the cylinder to the plane minus the origin. Around each of the curves, the vector field has a turning number: as you go around the curve counterclockwise, the vector field turns by some number of rotations (counting counterclockwise as positive. The index of the vector field in the compact subannulus is the difference: the number of turns on the outer boundary minus the number of turns on the inner boundary.

One way to describe a general formula is this: let $N^n$ be manifold, and let $M, \partial M \subset N$ be a compact submanifold. Let $X$ be a vector field that is is nonvanishing in a neighborhood of $\partial M$.

Choose an outward normal vector field $U$ along $\partial M$; now arrange $X$ so that its direction coincides with $U$ only in isolated points, so if we project $X$ to $N$ along $U$, it is a vector field with isolated singularities. Let $i+(X)$ be the sum of the Poincaré-Hopf indices over all singularities where $X$ is oriented outward. Then the Poincaré-Hopf index $i(X)$ of $X$ in $M$ equals the Euler characteristic of $M$ minus $i_+(X)$.

Here's one proof: triangulate a neighborhood of $N$ so that $\partial M$ and $T$ are subcomplexes, and so that $X$ is transverse to the triangulation except near the singularities, in the sense that in any simplex, the foliation defined by $X$ is topologically equivalent to the kernel of a linear map in general position of the simplex to $\mathbb R^{n-1}$. Put a $+1$ at the barycenter of each triangle of even dimension, and a $-1$ at the barycenter of each triangle of odd dimension. Think of $X$ as a wind that blows these numbers along, so that after an instant, all numbers (except for exceptions near the zeros of $X$) are inside an $n$-simplex. In any typical simplex, all the signs cancel out. However, along the boundary, some of the numbers are blown away and lost. To regularize the situation, modify $X$ by pushing in the negative normal direction. Now $X$ points inward everywhere except in a neighborhood of points where it coincides with the outward normal. Thus everything cancels out except for local contributions given by $i(X)$ and $i_+(X)$.

Every noncompact manifold admits nonzero vector fields, or more generally, vector fields with any specified set of isolated zeros along with the behavior near that zero.

However, if you have information of the behavior of a vector field near infinity, or just in a neighborhood of the boundary of a compact set, there is an index theorem. Perhaps this is the case with your $\mathbb T^! \times \mathbb R$.

In the particular case of a cylinder, there is a simple way to calculate the index. Take any compact subcylinder delimited by two circles. Map the cylinder to the plane minus the origin. Around each of the curves, the vector field has a turning number: as you go around the curve counterclockwise, the vector field turns by some number of rotations (counting counterclockwise as positive. The index of the vector field in the compact subannulus is the difference: the number of turns on the outer boundary minus the number of turns on the inner boundary.

One way to describe a general formula is this: let $N^n$ be manifold, and let $M, \partial M \subset N$ be a compact submanifold. Let $X$ be a vector field that is is nonvanishing in a neighborhood of $\partial M$.

Choose an outward normal vector field $U$ along $\partial M$; now arrange $X$ so that its direction coincides with $U$ only in isolated points, so if we project $X$ to $N$ along $U$, it is a vector field with isolated singularities. Let $i_+(X)$ be the sum of the Poincaré-Hopf indices over all singularities where $X$ is oriented outward. Then the Poincaré-Hopf index $i(X)$ of $X$ in $M$ equals the Euler characteristic of $M$ minus $i_+(X)$.

Here's one proof: triangulate a neighborhood of $M$ so that $\partial M$ is a subcomplex, and so that $X$ is transverse to the triangulation except near the singularities, in the sense that in any simplex, the foliation defined by $X$ is topologically equivalent to the kernel of a linear map in general position of the simplex to $\mathbb R^{n-1}$. Put a $+1$ at the barycenter of each triangle of even dimension, and a $-1$ at the barycenter of each triangle of odd dimension. Think of $X$ as a wind that blows these numbers along, so that after an instant, all numbers (except for exceptions near the zeros of $X$) are inside an $n$-simplex. In any typical simplex, all the signs cancel out. However, along the boundary, some of the numbers are blown away and lost. To regularize the situation, modify $X$ by pushing in the negative normal direction. Now $X$ points inward everywhere except in a neighborhood of points where it coincides with the outward normal. Thus everything cancels out except for local contributions given by $i(X)$ and $i_+(X)$.

Every noncompact manifold admits nonzero vector fields, or more generally, vector fields with any specified set of isolated zeros along with the behavior near that zero.

However, if you have information of the behavior of a vector field near infinity, or just in a neighborhood of the boundary of a compact set, there is an index theorem. Perhaps this is the case with your $\mathbb T^! \times \mathbb R$.

One formulation is this: let $N^n$ be manifold, and let $M, \partial M \subset N$ be a compact submanifold. Let $X$ be a vector field that is is nonvanishing in a neighborhood of $\partial M$. Suppose that the set of tangencies of $X$ to $\partial M$ consists of a codimension one submanifold $T$ of $\partial M$, so we can write $ \partial M$ as the union of two submanifolds with boundary, $\partial M = B_+ \cup B_-$ where the flow of $X$ is outward on $B_+$ and inward on $B_-$. Then the Poincaré-Hopf index for $X$ in $M$ is $\chi(M) - \chi(B_+)$

Here's one proof: triangulate a neighborhood of $N$ so that $\partial M$ and $T$ are subcomplexes, and so that $X$ is transverse to the triangulation except near the singularities, in the sense that in any simplex, the foliation defined by $X$ is topologically equivalent to the kernel of a linear map in general position of the simplex to $\mathbb R^{n-1}$. Put a $+1$ at the barycenter of each triangle of even dimension, and a $-1$ at the barycenter of each triangle of odd dimension. Think of $X$ as a wind that blows these numbers along, so that after an instant, all numbers (except for exceptions near the zeros of $X$) are inside an $n$-simplex. In any typical simplex, all the signs cancel out. However, the numbers on $B_+$ (which includes $T$) are blown away and lost. Near any zero of $X$ there is a mismatch, which can only depend on the local behavior of $X$. From the special case that $N$ is a closed manifold and $M = N$, it follows that is the Poincaré-Hopf index.

Every noncompact manifold admits nonzero vector fields, or more generally, vector fields with any specified set of isolated zeros along with the behavior near that zero.

However, if you have information of the behavior of a vector field near infinity, or just in a neighborhood of the boundary of a compact set, there is an index theorem. Perhaps this is the case with your $\mathbb T^! \times \mathbb R$.

In the particular case of a cylinder, there is a simple way to calculate the index. Take any compact subcylinder delimited by two circles. Map the cylinder to the plane minus the origin. Around each of the curves, the vector field has a turning number: as you go around the curve counterclockwise, the vector field turns by some number of rotations (counting counterclockwise as positive. The index of the vector field in the compact subannulus is the difference: the number of turns on the outer boundary minus the number of turns on the inner boundary.

One way to describe a general formula is this: let $N^n$ be manifold, and let $M, \partial M \subset N$ be a compact submanifold. Let $X$ be a vector field that is is nonvanishing in a neighborhood of $\partial M$.

Choose an outward normal vector field $U$ along $\partial M$; now arrange $X$ so that its direction coincides with $U$ only in isolated points, so if we project $X$ to $N$ along $U$, it is a vector field with isolated singularities. Let $i+(X)$ be the sum of the Poincaré-Hopf indices over all singularities where $X$ is oriented outward. Then the Poincaré-Hopf index $i(X)$ of $X$ in $M$ equals the Euler characteristic of $M$ minus $i_+(X)$.

Here's one proof: triangulate a neighborhood of $N$ so that $\partial M$ and $T$ are subcomplexes, and so that $X$ is transverse to the triangulation except near the singularities, in the sense that in any simplex, the foliation defined by $X$ is topologically equivalent to the kernel of a linear map in general position of the simplex to $\mathbb R^{n-1}$. Put a $+1$ at the barycenter of each triangle of even dimension, and a $-1$ at the barycenter of each triangle of odd dimension. Think of $X$ as a wind that blows these numbers along, so that after an instant, all numbers (except for exceptions near the zeros of $X$) are inside an $n$-simplex. In any typical simplex, all the signs cancel out. However, along the boundary, some of the numbers are blown away and lost. To regularize the situation, modify $X$ by pushing in the negative normal direction. Now $X$ points inward everywhere except in a neighborhood of points where it coincides with the outward normal. Thus everything cancels out except for local contributions given by $i(X)$ and $i_+(X)$.