Consider a vector bundle $V\to E\to X$ with fiber $V$, with structure group $G$, and $X$ path-connected. Consider a connection $\nabla$ on $E$. Then for any loop $L$ in $X$, based at $p$, we have a mapping: $$ hol: L\mapsto hol(L)\in Aut(V), $$

the holonomy map, which gives us the (linear) transformation of vectors after parallel transport around the loop $L$. Its image $Hol(\nabla)$ (dropping the reference to the base point) is a subgroup of $Aut(V)$.

If $X$ is simply connected, then it is known that $Hol(\nabla)$ is a path-connected Lie subgroup of $Aut(V)$. (See for example "Riemannian Holonomy Groups and Calibrated Geometry" by D. Joyce, chapter 2.)

This can be seen in the following way: a continuous homotopy $L_s, s\in[0,1]$ between two loops $L_0$ and $L_1$ defines a continuous curve in $Hol(\nabla)$, given by: $$ \gamma:s\mapsto \gamma_s= hol(L_s). $$

Therefore, homotopic loops have path-connected images in $Hol(\nabla)$.

My question is: how further can we go?

Between points in $Hol(\nabla)$, having a continuous curve that joins them is an equivalence relation. If we denote by $\pi_0(Hol(\nabla))$ the equivalence classes of such relation, we have a well defined map $hol^1:\pi_1(X)\to \pi_0(Hol(\nabla))$. In fact, if two loops are homotopic, their images lie in the same class.

$\pi_0$ has no group structure defined (would it be possible to define it?), so our map $hol^1$ is not a morphism of groups.

Now, a 2-dimensional homotopy $S^2\to X$ can be thought of a homotopy $L_s$ of loops, like before, but where both $L_1$ and $L_0$ are trivial (while for generic $s$, $L_s$ may not be trivial). The curve $s\mapsto hol(L_s)$ now defines a closed loop in $Hol(\nabla)$, based at the identity. A homotopy between two 2-dimensional homotopies in $X$ defines a homotopy of loops in $Hol(\nabla)$. In particular, we can define a mapping $hol^2:\pi_2(X)\to \pi_1(Hol(\nabla))$. It is well defined, because as we saw $hol$ preserves homotopy of different orders.

So in general, I would expect that $hol$ defines mappings: $$ hol^k:\pi_k(X)\to \pi_{k-1}(Hol(\nabla)). $$

Moreover, for $k>0$, $\pi_k$ are groups, and since composing homotopies of order $k$ in $X$ corresponds to composing homotopies of order $k-1$ in $Hol(\nabla)$ (and the image of trivial homotopies are trivial), the maps $hol^k$ are morphisms at least for $k>1$. So I would say:

The holonomy map defines morphisms $hol^k:\pi_k(X)\to \pi_{k-1}(Hol(\nabla))$.

This would imply:

A bundle $V\to E\to X$ with holonomy $Hol(\nabla)$ can exist only if every $\pi_k(Hol(\nabla))$ admits as a subgroup a homomorphic image (i.e. a quotient) of $\pi_{k+1}(M)$.

Is this reasoning correct? Is it helpful? Has it been done anywhere?

EDIT: For $TS^2$, the tangent bundle to the 2-sphere, and the Levi-Civita connection, the Gauss-Bonnet theorem can be stated in this (beautiful!) way: since $Hol(\nabla)=SO(2)\cong S^1$,

The morphism $hol^2: \pi_2(S^2) \to \pi_1(S^1)$ is an isomorphism.

Does this result generalize? Thank you!

Consider a vector bundle $V\to E\to X$ with fiber $V$, with structure group $G$, and $X$ path-connected. Consider a connection $\nabla$ on $E$. Then for any loop $L$ in $X$, based at $p$, we have a mapping: $$ hol: L\mapsto hol(L)\in Aut(V), $$

the holonomy map, which gives us the (linear) transformation of vectors after parallel transport around the loop $L$. Its image $Hol(\nabla)$ (dropping the reference to the base point) is a subgroup of $Aut(V)$.

If $X$ is simply connected, then it is known that $Hol(\nabla)$ is a path-connected Lie subgroup of $Aut(V)$. (See for example "Riemannian Holonomy Groups and Calibrated Geometry" by D. Joyce, chapter 2.)

This can be seen in the following way: a continuous homotopy $L_s, s\in[0,1]$ between two loops $L_0$ and $L_1$ defines a continuous curve in $Hol(\nabla)$, given by: $$ \gamma:s\mapsto \gamma_s= hol(L_s). $$

Therefore, homotopic loops have path-connected images in $Hol(\nabla)$.

My question is: how further can we go?

Between points in $Hol(\nabla)$, having a continuous curve that joins them is an equivalence relation. If we denote by $\pi_0(Hol(\nabla))$ the equivalence classes of such relation, we have a well defined map $hol^1:\pi_1(X)\to \pi_0(Hol(\nabla))$. In fact, if two loops are homotopic, their images lie in the same class.

$\pi_0$ has no group structure defined (would it be possible to define it?), so our map $hol^1$ is not a morphism of groups.

Now, a 2-dimensional homotopy $S^2\to X$ can be thought of a homotopy $L_s$ of loops, like before, but where both $L_1$ and $L_0$ are trivial (while for generic $s$, $L_s$ may not be trivial). The curve $s\mapsto hol(L_s)$ now defines a closed loop in $Hol(\nabla)$, based at the identity. A homotopy between two 2-dimensional homotopies in $X$ defines a homotopy of loops in $Hol(\nabla)$. In particular, we can define a mapping $hol^2:\pi_2(X)\to \pi_1(Hol(\nabla))$. It is well defined, because as we saw $hol$ preserves homotopy of different orders.

So in general, I would expect that $hol$ defines mappings: $$ hol^k:\pi_k(X)\to \pi_{k-1}(Hol(\nabla)). $$

Moreover, for $k>0$, $\pi_k$ are groups, and since composing homotopies of order $k$ in $X$ corresponds to composing homotopies of order $k-1$ in $Hol(\nabla)$ (and the image of trivial homotopies are trivial), the maps $hol^k$ are morphisms at least for $k>1$. So I would say:

The holonomy map defines morphisms $hol^k:\pi_k(X)\to \pi_{k-1}(Hol(\nabla))$.

This would imply:

A bundle $V\to E\to X$ with holonomy $Hol(\nabla)$ can exist only if every $\pi_k(Hol(\nabla))$ admits as a subgroup a homomorphic image (i.e. a quotient) of $\pi_{k+1}(M)$.

Is this reasoning correct? Is it helpful? Has it been done anywhere?

EDIT: For $TS^2$, the tangent bundle to the 2-sphere, and the Levi-Civita connection, the Gauss-Bonnet theorem can be stated in this (beautiful!) way: since $Hol(\nabla)=SO(2)\cong S^1$,

The morphism $hol^2: \pi_2(S^2) \to \pi_1(S^1)$ induces an isomorphism $\Bbb{Z}\to 2\Bbb{Z}$.

Does this result generalize? Thank you!

Consider a vector bundle $V\to E\to X$ with fiber $V$, with structure group $G$, and $X$ path-connected. Consider a connection $\nabla$ on $E$. Then for any loop $L$ in $X$, based at $p$, we have a mapping: $$ hol: L\mapsto hol(L)\in Aut(V), $$

the holonomy map, which gives us the (linear) transformation of vectors after parallel transport around the loop $L$. Its image $Hol(\nabla)$ (dropping the reference to the base point) is a subgroup of $Aut(V)$.

If $X$ is simply connected, then it is known that $Hol(\nabla)$ is a path-connected Lie subgroup of $Aut(V)$. (See for example "Riemannian Holonomy Groups and Calibrated Geometry" by D. Joyce, chapter 2.)

This can be seen in the following way: a continuous homotopy $L_s, s\in[0,1]$ between two loops $L_0$ and $L_1$ defines a continuous curve in $Hol(\nabla)$, given by: $$ \gamma:s\mapsto \gamma_s= hol(L_s). $$

Therefore, homotopic loops have path-connected images in $Hol(\nabla)$.

My question is: how further can we go?

Between points in $Hol(\nabla)$, having a continuous curve that joins them is an equivalence relation. If we denote by $\pi_0(Hol(\nabla))$ the equivalence classes of such relation, we have a well defined map $hol^1:\pi_1(X)\to \pi_0(Hol(\nabla))$. In fact, if two loops are homotopic, their images lie in the same class.

$\pi_0$ has no group structure defined (would it be possible to define it?), so out map $hol^1$ is not a morphism of groups.

Now, a 2-dimensional homotopy $S^2\to X$ can be thought of a homotopy $L_s$ of loops, like before, but where both $L_1$ and $L_0$ are trivial (while for generic $s$, $L_s$ may not be trivial). The curve $s\mapsto hol(L_s)$ now defines a closed loop in $Hol(\nabla)$, based at the identity. A homotopy between two 2-dimensional homotopies in $X$ defines a homotopy of loops in $Hol(\nabla)$. In particular, we can define a mapping $hol^2:\pi_2(X)\to \pi_1(Hol(\nabla))$. It is well defined, because as we saw $hol$ preserves homotopy of different orders.

So in general, I would expect that $hol$ defines mappings: $$ hol^k:\pi_k(X)\to \pi_{k-1}(Hol(\nabla)). $$

Moreover, for $k>0$, $\pi_k$ are groups, and since composing homotopies of order $k$ in $X$ corresponds to composing homotopies of order $k-1$ in $Hol(\nabla)$ (and the image of trivial homotopies are trivial), the maps $hol^k$ are morphisms at least for $k>1$. So I would say:

The holonomy map defines morphisms $hol^k:\pi_k(X)\to \pi_{k-1}(Hol(\nabla))$.

This would imply:

A bundle $V\to E\to X$ with holonomy $Hol(\nabla)$ can exist only if every $\pi_k(Hol(\nabla))$ admits as a subgroup a homomorphic image (i.e. a quotient) of $\pi_{k+1}(M)$.

Is this reasoning correct? Is it helpful? Has it been done anywhere?

EDIT: For $S^2$, its tangent bundle, and the Levi-Civita connection, the Gauss-Bonnet theorem can be stated in this (beautiful!) way: since $SO(2)=S^1$,

The morphism $hol^2: \pi_2(S^2) \to \pi_1(S^1)$ is an isomorphism.

Does this result generalize? Thank you!

Consider a vector bundle $V\to E\to X$ with fiber $V$, with structure group $G$, and $X$ path-connected. Consider a connection $\nabla$ on $E$. Then for any loop $L$ in $X$, based at $p$, we have a mapping: $$ hol: L\mapsto hol(L)\in Aut(V), $$

the holonomy map, which gives us the (linear) transformation of vectors after parallel transport around the loop $L$. Its image $Hol(\nabla)$ (dropping the reference to the base point) is a subgroup of $Aut(V)$.

If $X$ is simply connected, then it is known that $Hol(\nabla)$ is a path-connected Lie subgroup of $Aut(V)$. (See for example "Riemannian Holonomy Groups and Calibrated Geometry" by D. Joyce, chapter 2.)

This can be seen in the following way: a continuous homotopy $L_s, s\in[0,1]$ between two loops $L_0$ and $L_1$ defines a continuous curve in $Hol(\nabla)$, given by: $$ \gamma:s\mapsto \gamma_s= hol(L_s). $$

Therefore, homotopic loops have path-connected images in $Hol(\nabla)$.

My question is: how further can we go?

Between points in $Hol(\nabla)$, having a continuous curve that joins them is an equivalence relation. If we denote by $\pi_0(Hol(\nabla))$ the equivalence classes of such relation, we have a well defined map $hol^1:\pi_1(X)\to \pi_0(Hol(\nabla))$. In fact, if two loops are homotopic, their images lie in the same class.

$\pi_0$ has no group structure defined (would it be possible to define it?), so our map $hol^1$ is not a morphism of groups.

Now, a 2-dimensional homotopy $S^2\to X$ can be thought of a homotopy $L_s$ of loops, like before, but where both $L_1$ and $L_0$ are trivial (while for generic $s$, $L_s$ may not be trivial). The curve $s\mapsto hol(L_s)$ now defines a closed loop in $Hol(\nabla)$, based at the identity. A homotopy between two 2-dimensional homotopies in $X$ defines a homotopy of loops in $Hol(\nabla)$. In particular, we can define a mapping $hol^2:\pi_2(X)\to \pi_1(Hol(\nabla))$. It is well defined, because as we saw $hol$ preserves homotopy of different orders.

So in general, I would expect that $hol$ defines mappings: $$ hol^k:\pi_k(X)\to \pi_{k-1}(Hol(\nabla)). $$

Moreover, for $k>0$, $\pi_k$ are groups, and since composing homotopies of order $k$ in $X$ corresponds to composing homotopies of order $k-1$ in $Hol(\nabla)$ (and the image of trivial homotopies are trivial), the maps $hol^k$ are morphisms at least for $k>1$. So I would say:

The holonomy map defines morphisms $hol^k:\pi_k(X)\to \pi_{k-1}(Hol(\nabla))$.

This would imply:

A bundle $V\to E\to X$ with holonomy $Hol(\nabla)$ can exist only if every $\pi_k(Hol(\nabla))$ admits as a subgroup a homomorphic image (i.e. a quotient) of $\pi_{k+1}(M)$.

Is this reasoning correct? Is it helpful? Has it been done anywhere?

EDIT: For $TS^2$, the tangent bundle to the 2-sphere, and the Levi-Civita connection, the Gauss-Bonnet theorem can be stated in this (beautiful!) way: since $Hol(\nabla)=SO(2)\cong S^1$,

The morphism $hol^2: \pi_2(S^2) \to \pi_1(S^1)$ is an isomorphism.

Does this result generalize? Thank you!