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Consider $\nabla$ a connection in a vector bundle above a smooth manifold $M$.Consider a local frame $\sigma=(\sigma_1, \sigma_2,...,\sigma_m )$ on a contractible open set $U\subset M$ and calculate the curvature matrix $\Omega$ with respect to this frame.Take $\theta_k$ a frame of $2$ forms in $TU.$ Look at the matrices $S_{k}$(with real smooth functions as entries) defined by the equation $$ \Omega=\sum_k \theta_k S_k .$$

Does it follow from Ambrose-Singer that the matrices $S_k(p)$ span the Lie algebra of $Hol_p^o(\nabla)$ for $p \in U?$ If so and if the $S_k$ are simultaneously skew symmetrizable, then doesn't it follow that $Hol_p^o(\nabla)$ is a subgroup of $O(n)?$

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Your first question is a bit ambiguous. Are you asking whether, for each $p\in U$, the matrices $S_k(p)$ span the Lie algebra of $\mathrm{Hol}^0_p(\nabla)$ or are you asking whether, after taking the span of the union of the images of $S_k(p)$ over all $p\in U$, the result is the Lie algebra of $\mathrm{Hol}^0_p(\nabla)$?

The answer to the first of these two questions is clearly 'no' because you could easily have a point $p$ where all of the $S_k(p)$ vanish and yet the Lie group $\mathrm{Hol}^0_p(\nabla)$ has positive dimension.

Perhaps more surprising, the answer to second version of the question is 'no' as well. One can construct examples of metrics on $\mathbb{R}^4=U=M$ together with a framing $\sigma$ as you describe such that the resulting matrices $S_k$ all take values in $\mathfrak{u}(2)\subset\mathfrak{so}(4)$, but the holonomy of the metric is equal to $\mathrm{SO}(4)$.

The answer to your second question is also 'no'. Here's an example: Let $M=\mathbb{R}^2$ (i.e., the $xy$-plane) and let the vector bundle be the trivial bundle $V = M\times\mathbb{R}^2$. Let $\nabla$ be the connection on $V$ whose connection matrix $\omega$ with respect to the standard basis $\sigma = (\sigma_1,\sigma_2$) of section is $$ \omega = \begin{pmatrix}\mathrm{d}x & \mathrm{d}y\\ \mathrm{d}y & -\mathrm{d}x\end{pmatrix}.\tag0 $$ Then the corresponding curvature matrix is $$ \Omega = \mathrm{d}\omega + \omega\wedge\omega = \begin{pmatrix}0& 2\,\mathrm{d}x\wedge\mathrm{d}y\\ -2\,\mathrm{d}x\wedge\mathrm{d}y & 0\end{pmatrix}, $$ so it takes values in $\mathfrak{so}(2)$. However, I claim that the holonomy of $\nabla$ is $\mathrm{SL}(2,\mathbb{R})$. To see this, note that the Lie algebra of the holonomy group must contain $\mathfrak{so}(2)$. Since there is no Lie algebra properly between $\mathfrak{so}(2)$ and $\mathfrak{sl}(2,\mathbb{R})$, it suffices to show that the holonomy of $\nabla$ is not $\mathrm{SO}(2)$.

Suppose that it were. Then there would exist a mapping $g:M\to \mathrm{GL}(2,\mathbb{R})$ and a $1$-form $\alpha$ such that $$ \omega = g^{-1}\,\mathrm{d}g + g^{-1}\begin{pmatrix}0&\alpha\\-\alpha&0\end{pmatrix}g.\tag1 $$ However, this would imply, by the standard calculation, that $$ \Omega = g^{-1}\begin{pmatrix}0&\mathrm{d}\alpha\\-\mathrm{d}\alpha&0\end{pmatrix}g. $$ Consequently, we would have to have $$ g^{-1}\begin{pmatrix}0&1\\-1&0\end{pmatrix}g = \begin{pmatrix}0&f\\-f&0\end{pmatrix} $$ for some function $f$ and, taking determinants, we find that $f^2 = 1$, so $f\equiv\pm 1$. This then implies, by algebra, that $g$ is of the form $$ g = \begin{pmatrix}u&v\\\mp v&\pm u\end{pmatrix} $$ for some functions $u$ and $v$ on $M$ that do not simultaneously vanish. Plugging this into the supposed formula (1) for $\omega$, we get that $\omega$ must be of the form $$ \omega = \begin{pmatrix}\mu &\nu\\-\nu&\mu\end{pmatrix} $$ for some $1$-forms $\mu$ and $\nu$. But this obviously contradicts the definition (0) of $\omega$.

Hence, the holonomy of $\nabla$, which is a connected Lie group properly containing $\mathrm{SO}(2)$ and contained in $\mathrm{SL}(2,\mathbb{R})$, must be equal to $\mathrm{SL}(2,\mathbb{R})$.

Added example: Let $M=\mathbb{R}^3$ with standard coordinates $(x^1,x^2,x^3)$, let the vector bundle be $V = TM$, and let $\sigma=(\sigma_1,\sigma_2,\sigma_3)$ where $\sigma_i$ is the $i$th-coordinate vector field. Let $\nabla$ be the connection that satisifies $\nabla_{\sigma_i}\sigma_i = 0$ and $\nabla_{\sigma_j}\sigma_i = \sigma_k$ whenever $i$, $j$, and $k$ are distinct. Then $\nabla$ is a torsion-free connection on $V=TM$ and the curvature matrix $\Omega$ is skew-symmetric, with the $S_k(p)$ spanning $\mathfrak{so}(3)$ at every point $p\in M$. Nevertheless, the holonomy of $\nabla$ is $\mathrm{SL}(3,\mathbb{R})$, not $\mathrm{SO}(3)$. This follows from an argument very similar to the one given above above.

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  • $\begingroup$ I apologize for the ambiguity. Yes my question was if the matrices $S_k(p)$ span the linear algebra of $Hol_p^o(\nabla)$ for all $p \in U.$ Your example answers my question. Thank you! $\endgroup$
    – Mike Cocos
    Dec 13, 2019 at 16:25
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    $\begingroup$ @MikeCocos: You are welcome. Is there something about my answer that you don't believe or understand? If so, let me know, and I'll try to answer your questions. I realize that many people misunderstand the Ambrose-Singer theorem, and your questions seem to stem from one of the common misunderstandings that is often not addressed in treatments of the subject. It's worthwhile to have these kinds of examples out there to help dispel these misconceptions. $\endgroup$ Dec 15, 2019 at 8:02
  • $\begingroup$ Professor Bryant. No there is no issue with your answer. You are always very helpful and to the point. It is my mind that works really slow. Besides not having enough time to spend on research my mind doesn't help much either. It is clear now to me that the curvature at a single point doesn't contain as much information as the restricted holonomy group. I might not need that much though since all I am looking for is a simple algorithm to determine when a connection is locally metric. Thank you so much for your kind help. $\endgroup$
    – Mike Cocos
    Dec 16, 2019 at 18:15
  • $\begingroup$ Please let me know if I can ask you for some input on a related problem. I am sure it wouldn't take too long for you to spot eventual mistakes but I don't want to take advantage of too much of your time. It is about affinely flat manifolds. Thank you for al your help! $\endgroup$
    – Mike Cocos
    Jan 23, 2020 at 16:41
  • $\begingroup$ @MikeCocos: Sure. I'd be happy to help if I can. You can email me directly if you don't want to use MO as an intermediary. $\endgroup$ Jan 23, 2020 at 17:28

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