Tensor fields defining $G$-structure are parallel

Question

Suppose $G \leq GL_n(\mathbb{R})$ is the stabilizer of some tensors $T^0_1, ..., T^0_k$, let $P$ be a $G$-structure on a manifold, i.e. a principal $G$ subbundle of the frame bundle of $M$ and let $T_1, ..., T_k$ be tensor fields that are pointwise the image of $T^0_1, ..., T^0_k$ through the frames of $P$, as in this question. Is it true that the tensor fields $T_1, ..., T_k$ are parallel with respect to any given connection on $P$? If you want add the hypothesis of the connection on $P$ being torsion-free (meaning the induced connection on the tangent bundle is) but I believe this is not needed. Alternatively, is it true that if the holonomy of a Riemannian manifold is contained in $G$ then the tensors defining (in the sense of the linked question) the $G$ structure are parallel? By looking at Berger's classification it looks true.

I guess that one way to go would be to argue that the parallel transport of a generic tensor on $M$ is characterized by the principal bundle this way: take a tensor at a point, pull it back through a frame at that point getting a tensor on $\mathbb{R}^n$, transport the frame to the point you want and finally push forward the tensor through the new frame. However I am not at all familiar with parallel transport on induced vector bundles to argue that this is true.

Answer 1

You are right, then tensor fields are indeed parallel and this is independent of torsion-freeness. In fact, if you take any $G$-invariant tensor $T_0\in\otimes^k\mathbb R^{n*}\otimes\otimes^\ell\mathbb R^n$, this gives rise to a $\binom\ell k$-tensor field $T$ on any manifold $M$ endowed with a $G$-structure $P$ by "pulling back along the frames in $P$". Now you can view the tensor bundle $\otimes^k T^*M\otimes\otimes^\ell TM$ as the associated bundle $P\times_G(\otimes^k\mathbb R^{n*}\otimes\otimes^\ell\mathbb R^n)$ and correspondingly, $\binom\ell k$-tensor fields on $M$ are in bijective correspondence with $G$-equivariant smooth functions $P\to \otimes^k\mathbb R^{n*}\otimes\otimes^\ell\mathbb R^n$. By construction, in this picture $T$ corresponds to the constant function $T_0$. But for any principal connection $\gamma$ on $P$ the induced connection $\nabla$ on $\binom\ell k$-tensor fields has the property that for a vector field $\xi\in\mathfrak X(M)$ the equivariant function corresponding to $\nabla_\xi T$ is obtained by differentiating the equivariant function corresponding to $T$ with respect to the horizontal lift of $\xi$ to $P$. But of course, if you differentiate a constant function, you always get $0$.

Answer 2

Remark1. Recall that the holonomy of a metric is the holonomy of its Levi-Civita connection. Thus a manifold may admit an affine connection with holonomy in $G<GL(n)$ and still not admitting any metric with holonomy in $G$.

Remark2. If there is a reduction of the structure group to $G<GL(n)$, then there is a connection with holonomy $G$. This is essentially the fact that any vector bundle admits a connection applied to the vector bundle induced by the reduction.

Prop1. Exists an $n$-manifolds $X$ and subgroup $G<GL(n), G = Stab(T)$ ($T$ in the tensor algebra of $\mathbb R^n$) such that the frame bundle of $X$ admits a reduction of the structure group to $G$ but $X$ does not admit a metric with holonomy in $G$. In particular there is no Riemannian connection on $TX$ which is torsion free and for which the tensor induced by $G$ is parallel.

proof: For example, an almost Hermitian manifold is a manifold with an almost complex structure $J$ and a compatible Riemannian metric $g$. The pair $(J,g)$ exists iff there is a reduction of the structure group to $U(2n)$. This condition is homotopy-theoretical in nature. Indeed by a theorem of Wu, we know that in dimension 4 such a structure exists iff the third integral Stiefel-Whitney class $W_3\in H^3(X,\mathbb Z)$ and $c^2-2\chi(M)-3\sigma(X)\in H^4(X,\mathbb Z)$ vanish ($c$ is the first Chern-class of $J$). However the existence of a torsion free connection for which $(J,g)$ is parallel would imply that the Nijenhuis tensor vanishes and hence $X$ would be a complex manifold with $J$ as induced ACS. This is not true in general. For example $\mathbb{CP}^2\#\mathbb{CP}^2\#\mathbb{CP}^2$ admits an ACS but not a a complex structure (a proof that uses the Bogomolov-Miyaoka-Yau may be found in these notes by Aleksandar Milivojević).

Prop2. Let $G = \bigcap_i Stab(T_i^0)< GL(n)$, $T_i^0$ in the tensor algebra of $\mathbb R^n$ and suppose that $X$ has a reduction to of the structure group to $G$ with associated $G$-principal bundle $P\to X$. Let $\nabla$ be a connection on $P$. Then the induced tensor fields $T_i$ obtained from $T^0_i$ are $\nabla$-parallel.

Proof: Since $\nabla$ is a connection $P$ it has holonomy contained in $G$. The tensors $T_i$ associated to the reduction are obtained by parallel transporting the model $T^0_1, \dots, T^0_k$ written in a G-frame given by the reduction. The holonomy contained in $G = Stab(T^0_k)$ ensures that this procedure yields well defined tensors $T_i$.

EDIT after comment. Consider a frame $F_{x_0}$ at $x_0 \in X$ in the reduction $G_{x_0}$ and consider a tensor $T_{i,x_0}$ at $x_0$ that has components equal to $T^0_i$ in this frame. Now define a tensor field $T_i$ by parallel transporting $T_{i,x_0}$. Consider a point $x\in X$. Then, since $T_i$ is defined by parallel transport, $T_i(x)$ will have components equal to $T^0_i$ in any frame obtained by parallel transporting $F_{x_0}$ to $x$ along any curve $\gamma$ joining $x_0$ to $x$. But since $G$ stabilizes $T^0_i$, if we represent $T_i(x)$ in any other frame in $G_x$ we will obtain the same components $T^0_i$.

Therefore the two definitions coincide.