[proofplan] The two directions are dual constructions linking parallel tensor fields on $M$ to algebraic invariants at one point. Forward, a parallel tensor field $\alpha$ satisfies $P_\gamma \alpha_x = \alpha_x$ for every loop $\gamma$ based at $x$ (parallel transport preserves a $\nabla$-parallel section), so the value $\alpha_0 := \alpha_x$ is fixed by the natural action of $\mathrm{Hol}_x(M)$ on the tensor space at $x$. Reverse, given $\alpha_0$ invariant under $\mathrm{Hol}_x(M)$, we define $\alpha$ at any $y \in M$ by parallel transporting $\alpha_0$ along any piecewise-smooth path $\sigma$ from $x$ to $y$; the value is independent of $\sigma$ precisely because two such paths differ by a loop at $x$, and Hol-invariance ensures the two transports agree. The construction is smooth, parallel, and recovers $\alpha_0$ at $x$. [/proofplan]

[step:Set up the action of $\mathrm{Hol}_x(M)$ on the tensor space at $x$]Let $(M, g)$ be a connected Riemannian manifold of dimension $n$, $\nabla$ its Levi-Civita connection, and fix $x \in M$. The holonomy group at $x$ is \begin{align*} \mathrm{Hol}_x(M) := \{P_\gamma : \gamma \in \Omega(x, x)\} \subseteq \mathrm{O}(T_x M, g_x), \end{align*} where $\Omega(x, x)$ denotes the space of piecewise-smooth loops based at $x$ and $P_\gamma : T_x M \to T_x M$ denotes parallel transport along $\gamma$ (see [Parallel Transport, Holonomy and Restricted Holonomy](/theorems/2762)). Containment in the orthogonal group follows from metric compatibility of the Levi-Civita connection: $g_x(P_\gamma v, P_\gamma w) = g_x(v, w)$ for all $v, w \in T_x M$. The natural action of $\mathrm{GL}(T_x M)$ on the tensor space \begin{align*} \mathcal{T}_x^{(p,q)} := T_x M^{\otimes p} \otimes T_x^* M^{\otimes q} \end{align*} is given on decomposable tensors by \begin{align*} A : \mathcal{T}_x^{(p,q)} &\to \mathcal{T}_x^{(p,q)}, \\ v_1 \otimes \cdots \otimes v_p \otimes \xi^1 \otimes \cdots \otimes \xi^q &\mapsto (Av_1) \otimes \cdots \otimes (Av_p) \otimes (\xi^1 \circ A^{-1}) \otimes \cdots \otimes (\xi^q \circ A^{-1}), \end{align*} extended by linearity. Under parallel transport $A = P_\gamma$, this is the natural lift of $P_\gamma$ to tensors of type $(p, q)$, and we denote it again by $P_\gamma : \mathcal{T}_x^{(p,q)} \to \mathcal{T}_x^{(p,q)}$. Restricting to $\mathrm{Hol}_x(M)$, an element $\alpha_0 \in \mathcal{T}_x^{(p,q)}$ is **Hol-invariant** if $P_\gamma \alpha_0 = \alpha_0$ for every $\gamma \in \Omega(x, x)$. The compatibility we need is that this lifted parallel transport on tensors agrees with parallel transport of tensor fields along $\gamma$ in the sense of the induced connection: if $T$ is a $(p,q)$-tensor field along $\gamma$ with $\nabla_{\dot\gamma} T = 0$ and $T(0) = \alpha_0$, then $T(1) = P_\gamma \alpha_0$ in $\mathcal{T}_x^{(p,q)}$. This is the definition of the induced parallel transport on tensor bundles, and it is a standard consequence of the Leibniz rule for $\nabla$ on tensors.[/step]

[guided]Before stating the equivalence, we need to make sense of the phrase "Hol-invariant tensor" — that is, we need to say how the holonomy group, which is a priori a subgroup of $\mathrm{GL}(T_x M)$, acts on the larger tensor space $T_x M^{\otimes p} \otimes T_x^* M^{\otimes q}$. We do this in three pieces: define the holonomy group, define the natural tensor-product action of $\mathrm{GL}(T_x M)$ on tensors, and verify that this lifted action matches induced parallel transport on tensor fields. Let $(M, g)$ be a connected Riemannian manifold of dimension $n$, $\nabla$ its Levi-Civita connection, and fix $x \in M$. The holonomy group at $x$ is built from parallel transport around loops: \begin{align*} \mathrm{Hol}_x(M) := \{P_\gamma : \gamma \in \Omega(x, x)\} \subseteq \mathrm{O}(T_x M, g_x), \end{align*} where $\Omega(x, x)$ is the space of piecewise-smooth loops based at $x$ and $P_\gamma : T_x M \to T_x M$ is parallel transport along $\gamma$ (see [Parallel Transport, Holonomy and Restricted Holonomy](/theorems/2762)). Why does this sit inside the orthogonal group? Because the Levi-Civita connection is metric-compatible, $\nabla g = 0$. Applying this to the parallel-transport ODE for the inner product gives $\frac{d}{dt} g(V(t), W(t)) = g(\nabla_{\dot\gamma} V, W) + g(V, \nabla_{\dot\gamma} W) = 0$ for $V, W$ parallel along $\gamma$, so $g(P_\gamma v, P_\gamma w) = g(v, w)$ — parallel transport preserves the metric. Next, we describe the natural action of $\mathrm{GL}(T_x M)$ on tensors. Set \begin{align*} \mathcal{T}_x^{(p,q)} := T_x M^{\otimes p} \otimes T_x^* M^{\otimes q}. \end{align*} A linear map $A \in \mathrm{GL}(T_x M)$ extends uniquely to $\mathcal{T}_x^{(p,q)}$ by the rule "act on each $V$-factor by $A$, on each $V^*$-factor by $(A^{-1})^*$": \begin{align*} A : \mathcal{T}_x^{(p,q)} &\to \mathcal{T}_x^{(p,q)}, \\ v_1 \otimes \cdots \otimes v_p \otimes \xi^1 \otimes \cdots \otimes \xi^q &\mapsto (Av_1) \otimes \cdots \otimes (Av_p) \otimes (\xi^1 \circ A^{-1}) \otimes \cdots \otimes (\xi^q \circ A^{-1}), \end{align*} extended by linearity. Why precompose covectors by $A^{-1}$ rather than just $A$? Because we want the canonical evaluation pairing $V^* \times V \to \mathbb{R}$ to be invariant: requiring $(A\xi)(Av) = \xi(v)$ forces $A\xi = \xi \circ A^{-1}$. This extension is functorial — it preserves composition and inverses — so it defines a homomorphism $\mathrm{GL}(T_x M) \to \mathrm{GL}(\mathcal{T}_x^{(p,q)})$. For $A = P_\gamma$ this gives the lifted parallel transport on tensors, which we again denote $P_\gamma : \mathcal{T}_x^{(p,q)} \to \mathcal{T}_x^{(p,q)}$. We say $\alpha_0 \in \mathcal{T}_x^{(p,q)}$ is **Hol-invariant** if $P_\gamma \alpha_0 = \alpha_0$ for every $\gamma \in \Omega(x, x)$. Finally, we need a compatibility: the lifted $P_\gamma$ on the algebraic tensor space agrees with parallel transport of tensor fields along $\gamma$ in the sense of the induced connection on $\mathcal{T}^{(p,q)} M$. Concretely, if $T$ is a $(p,q)$-tensor field along $\gamma$ with $\nabla_{\dot\gamma} T = 0$ and $T(0) = \alpha_0$, then $T(1) = P_\gamma \alpha_0$. Why is this true? The induced connection on tensors is defined precisely by the Leibniz rule: $\nabla(X \otimes Y) = (\nabla X) \otimes Y + X \otimes \nabla Y$, and dually $(\nabla\xi)(Y) = \nabla(\xi(Y)) - \xi(\nabla Y)$. Solving $\nabla_{\dot\gamma} T = 0$ along $\gamma$ for a tensor with initial value $\alpha_0$ produces the same answer at $t = 1$ as applying the lifted $P_\gamma$ to $\alpha_0$ — both are the unique linear lift consistent with the Leibniz rule. This compatibility is what allows us to translate fluently between "parallel tensor field" (an analytic object) and "fixed point of $\mathrm{Hol}_x(M)$" (an algebraic object) in the next steps.[/guided]

[step:Forward direction: a parallel tensor field gives a Hol-invariant tensor at $x$] Suppose $\alpha$ is a $(p, q)$-tensor field on $M$ with $\nabla \alpha = 0$. Set $\alpha_0 := \alpha_x \in \mathcal{T}_x^{(p,q)}$. We claim $\alpha_0$ is Hol-invariant. Fix $\gamma \in \Omega(x, x)$. Restrict $\alpha$ to $\gamma$: the tensor field along $\gamma$ given by $T(t) := \alpha_{\gamma(t)}$ satisfies \begin{align*} \nabla_{\dot\gamma(t)} T(t) = (\nabla \alpha)_{\gamma(t)}(\dot\gamma(t)) = 0 \end{align*} because $\nabla \alpha \equiv 0$ as a section of $\mathcal{T}^{(p,q+1)}M$. Hence $T$ is parallel along $\gamma$ with $T(0) = \alpha_x = \alpha_0$. By definition of the induced parallel transport on tensors, $T(1) = P_\gamma \alpha_0$. But also $T(1) = \alpha_{\gamma(1)} = \alpha_x = \alpha_0$ since $\gamma$ is a loop. Combining, \begin{align*} P_\gamma \alpha_0 = \alpha_0. \end{align*} Since $\gamma$ was an arbitrary loop at $x$, this holds for every element of $\mathrm{Hol}_x(M)$, so $\alpha_0$ is Hol-invariant. [/step]

[step:Reverse direction: define $\alpha$ globally by parallel transport from $x$]Conversely, suppose $\alpha_0 \in \mathcal{T}_x^{(p,q)}$ is invariant under $\mathrm{Hol}_x(M)$. We construct a $(p, q)$-tensor field $\alpha$ on $M$ with $\alpha_x = \alpha_0$ and $\nabla \alpha = 0$. Fix $y \in M$. Since $M$ is connected (and connected smooth manifolds are path-connected), there exists a piecewise-smooth path $\sigma : [0, 1] \to M$ with $\sigma(0) = x$, $\sigma(1) = y$. Define \begin{align*} \alpha_y := P_\sigma \alpha_0 \in \mathcal{T}_y^{(p,q)}, \end{align*} where $P_\sigma : \mathcal{T}_x^{(p,q)} \to \mathcal{T}_y^{(p,q)}$ is the induced parallel transport on tensor spaces. [claim:The value $\alpha_y$ is independent of the choice of path $\sigma$ from $x$ to $y$] [proof] Let $\sigma_1, \sigma_2$ be two piecewise-smooth paths from $x$ to $y$. Form the loop $\gamma := \sigma_2^{-1} * \sigma_1$ at $x$ (concatenation: traverse $\sigma_1$ from $x$ to $y$, then $\sigma_2$ in reverse from $y$ back to $x$). This $\gamma$ is a piecewise-smooth loop at $x$. Parallel transport is functorial under concatenation and reversal: $P_{\sigma_2^{-1}} = (P_{\sigma_2})^{-1}$ and $P_{\sigma_2^{-1} * \sigma_1} = P_{\sigma_2^{-1}} \circ P_{\sigma_1} = (P_{\sigma_2})^{-1} \circ P_{\sigma_1}$. Hence \begin{align*} P_\gamma = (P_{\sigma_2})^{-1} \circ P_{\sigma_1} : \mathcal{T}_x^{(p,q)} \to \mathcal{T}_x^{(p,q)}. \end{align*} The element $P_\gamma$ lies in (the tensor lift of) $\mathrm{Hol}_x(M)$, so by Hol-invariance of $\alpha_0$, \begin{align*} P_\gamma \alpha_0 = \alpha_0 \iff (P_{\sigma_2})^{-1}(P_{\sigma_1} \alpha_0) = \alpha_0 \iff P_{\sigma_1} \alpha_0 = P_{\sigma_2} \alpha_0. \end{align*} This is the desired equality of $\alpha_y$ computed from $\sigma_1$ and from $\sigma_2$. [/proof] [/claim] The well-definedness claim shows $\alpha : M \to \mathcal{T}^{(p,q)} M$, $y \mapsto \alpha_y$, is a well-defined section of the tensor bundle. By construction, $\alpha_x = P_{c_x} \alpha_0 = \alpha_0$ (parallel transport along the constant loop is the identity).[/step]

[guided]The construction of $\alpha$ is the dual of the forward direction. To define $\alpha$ at a point $y \neq x$, we have essentially one rule available: parallel-transport-from-$x$. Connectedness gives at least one path $\sigma$ from $x$ to $y$ — recall that connected smooth manifolds are path-connected — so we set \begin{align*} \alpha_y := P_\sigma \alpha_0 \in \mathcal{T}_y^{(p,q)}, \end{align*} where $P_\sigma : \mathcal{T}_x^{(p,q)} \to \mathcal{T}_y^{(p,q)}$ is the induced parallel transport on tensor spaces. The hard question is: what if there is more than one path? Then $P_{\sigma_1} \alpha_0$ and $P_{\sigma_2} \alpha_0$ are a priori two different elements of $\mathcal{T}_y^{(p,q)}$, and we need to check they agree. Take two piecewise-smooth paths $\sigma_1, \sigma_2$ from $x$ to $y$. The trick is to convert the path-independence question into a holonomy question by closing up: form the loop $\gamma := \sigma_2^{-1} * \sigma_1$ at $x$ (traverse $\sigma_1$ forward from $x$ to $y$, then $\sigma_2$ in reverse from $y$ back to $x$). This is again piecewise-smooth, so $\gamma \in \Omega(x, x)$. Parallel transport is functorial under concatenation and reversal — $P_{\sigma_2^{-1}} = (P_{\sigma_2})^{-1}$ and $P_{\sigma_2^{-1} * \sigma_1} = P_{\sigma_2^{-1}} \circ P_{\sigma_1}$ — so \begin{align*} P_\gamma = (P_{\sigma_2})^{-1} \circ P_{\sigma_1} : \mathcal{T}_x^{(p,q)} \to \mathcal{T}_x^{(p,q)}. \end{align*} Now $P_\gamma$ (in its lifted action on tensors) lies in $\mathrm{Hol}_x(M)$, so by hypothesis $\alpha_0$ is fixed by it: \begin{align*} P_\gamma \alpha_0 = \alpha_0 \iff (P_{\sigma_2})^{-1}(P_{\sigma_1} \alpha_0) = \alpha_0 \iff P_{\sigma_1} \alpha_0 = P_{\sigma_2} \alpha_0. \end{align*} This is exactly the path-independence we wanted: $\alpha_y$ is well-defined. Notice the role of Hol-invariance here. If $\alpha_0$ were *not* fixed by all elements of $\mathrm{Hol}_x(M)$, then for some pair $\sigma_1, \sigma_2$ we would have $P_{\sigma_1} \alpha_0 \neq P_{\sigma_2} \alpha_0$, and the construction would fail to give a single value at $y$. This is why the equivalence is exact and not just an implication: the reverse direction *requires* Hol-invariance of $\alpha_0$ as the precise obstruction to path-independence. The construction works for an algebraic invariant of the holonomy group; nothing weaker suffices. The well-definedness shows $\alpha : M \to \mathcal{T}^{(p,q)} M$, $y \mapsto \alpha_y$, is a well-defined section of the tensor bundle. Evaluating at $x$, the constant loop $c_x$ gives identity parallel transport, so $\alpha_x = P_{c_x} \alpha_0 = \alpha_0$ as required. The connectedness hypothesis on $M$ is essential at this step: if $M$ had two components, an invariant $\alpha_0$ at $x$ would only define $\alpha$ on the component of $x$, and there would be no way to fix $\alpha$ on the other component. The theorem's restriction to connected manifolds is therefore not cosmetic — it is forced by the construction.[/guided]

[step:Show that the constructed $\alpha$ is smooth]We verify that $y \mapsto \alpha_y$ is smooth as a section of $\mathcal{T}^{(p,q)} M$. Let $y_0 \in M$ and let $U$ be a normal neighbourhood of $y_0$, i.e., $U = \exp_{y_0}(V)$ for an open star-shaped $V \subseteq T_{y_0} M$ on which $\exp_{y_0}$ is a diffeomorphism. Fix a piecewise-smooth path $\sigma_0$ from $x$ to $y_0$. For each $y \in U$, set $\sigma_y$ to be the concatenation of $\sigma_0$ followed by the radial geodesic $t \mapsto \exp_{y_0}(t \exp_{y_0}^{-1}(y))$ for $t \in [0, 1]$. Then $\sigma_y$ is a piecewise-smooth path from $x$ to $y$, and \begin{align*} \alpha_y = P_{\sigma_y} \alpha_0 = P_{\mathrm{rad}, y_0 \to y} \circ P_{\sigma_0} \alpha_0 = P_{\mathrm{rad}, y_0 \to y} \alpha_{y_0}, \end{align*} where $P_{\mathrm{rad}, y_0 \to y}$ is parallel transport along the radial geodesic from $y_0$ to $y$ inside $U$. The map $y \mapsto P_{\mathrm{rad}, y_0 \to y}$ is smooth from $U$ into the bundle of linear isomorphisms of tensor spaces — this is the standard smooth dependence of solutions of an ODE (the parallel transport equation) on initial conditions and parameters, applied to the smooth radial-geodesic family. Hence $y \mapsto \alpha_y = P_{\mathrm{rad}, y_0 \to y} \alpha_{y_0}$ is smooth on $U$. Since $y_0$ was arbitrary, $\alpha$ is smooth on $M$.[/step]

[guided]We need to show $y \mapsto \alpha_y$ is smooth as a section of $\mathcal{T}^{(p,q)} M$. The naive approach — read off smoothness from the formula $\alpha_y = P_{\sigma_y}\alpha_0$ — is unworkable, because the path $\sigma_y$ from $x$ to $y$ varies discontinuously as $y$ moves around the manifold. The fix is to work locally and exploit the well-definedness from Step 3: since $\alpha_y$ does not depend on the choice of path, we are free to pick the most convenient family of paths near each point. Fix $y_0 \in M$. Take a normal neighbourhood $U$ of $y_0$, i.e., $U = \exp_{y_0}(V)$ for an open star-shaped $V \subseteq T_{y_0} M$ on which $\exp_{y_0}$ is a diffeomorphism. Why normal neighbourhoods? Because they come equipped with a canonical smooth family of paths from $y_0$ to nearby points — the radial geodesics $t \mapsto \exp_{y_0}(t \exp_{y_0}^{-1}(y))$ for $t \in [0,1]$ — and these depend smoothly on $y \in U$. Fix any piecewise-smooth path $\sigma_0$ from $x$ to $y_0$ (this choice is fixed once and for all, independent of $y$). For each $y \in U$, let $\sigma_y$ be the concatenation of $\sigma_0$ followed by the radial geodesic from $y_0$ to $y$. Then $\sigma_y$ is a piecewise-smooth path from $x$ to $y$, and by the well-definedness in Step 3 we may compute $\alpha_y$ using $\sigma_y$: \begin{align*} \alpha_y = P_{\sigma_y} \alpha_0 = P_{\mathrm{rad}, y_0 \to y} \circ P_{\sigma_0} \alpha_0 = P_{\mathrm{rad}, y_0 \to y} \alpha_{y_0}, \end{align*} where $P_{\mathrm{rad}, y_0 \to y}$ is parallel transport along the radial geodesic from $y_0$ to $y$ inside $U$. The factorisation uses functoriality of parallel transport under concatenation, and the last equality uses $\alpha_{y_0} = P_{\sigma_0} \alpha_0$ (the well-definedness lets us evaluate $\alpha_{y_0}$ via $\sigma_0$). Now the smoothness is immediate. The first factor $P_{\sigma_0}$ contributes the constant $\alpha_{y_0} \in \mathcal{T}_{y_0}^{(p,q)}$ — no $y$-dependence at all. The remaining map $y \mapsto P_{\mathrm{rad}, y_0 \to y}$ is smooth from $U$ into the bundle of linear isomorphisms of tensor spaces: this is the standard smooth-dependence-on-data theorem for ODEs, applied to the parallel-transport equation along the smoothly-varying radial-geodesic family. Composing with the fixed vector $\alpha_{y_0}$ preserves smoothness, so $y \mapsto \alpha_y$ is smooth on $U$. Since $y_0$ was arbitrary and smoothness is a local property, $\alpha$ is smooth on $M$. The conceptual takeaway: well-definedness from Step 3 is what makes the local smoothness argument possible. Without it, we would be stuck with the discontinuous global path $\sigma_y$; with it, we can swap to the smooth radial paths in each chart, and the chart-by-chart smoothness assembles to global smoothness.[/guided]

[step:Show that the constructed $\alpha$ is parallel: $\nabla \alpha = 0$]We verify $\nabla \alpha = 0$. Fix $y \in M$ and a tangent vector $v \in T_y M$. Choose any smooth curve $\eta : (-\varepsilon, \varepsilon) \to M$ with $\eta(0) = y$, $\dot\eta(0) = v$. For each $s \in (-\varepsilon, \varepsilon)$, let $\sigma_s$ be a path from $x$ to $\eta(s)$ obtained by concatenating a fixed path $\sigma_0$ from $x$ to $y$ with the curve segment $\eta|_{[0, s]}$ (read as a path from $y$ to $\eta(s)$). Then \begin{align*} \alpha_{\eta(s)} = P_{\sigma_s} \alpha_0 = P_{\eta|_{[0,s]}} \circ P_{\sigma_0} \alpha_0 = P_{\eta|_{[0,s]}} \alpha_y. \end{align*} The $s$-derivative of $\alpha_{\eta(s)}$ at $s = 0$ is, by the very definition of $\nabla_v \alpha$ as the rate of change of a section measured against parallel transport, \begin{align*} (\nabla_v \alpha)(y) = \lim_{s \to 0} \frac{1}{s} \big[ (P_{\eta|_{[0,s]}})^{-1} \alpha_{\eta(s)} - \alpha_y \big] = \lim_{s \to 0} \frac{1}{s} \big[ \alpha_y - \alpha_y \big] = 0. \end{align*} Since $v$ and $y$ were arbitrary, $\nabla \alpha = 0$ on all of $M$.[/step]

[guided]We now check $\nabla \alpha = 0$. The verification is essentially tautological — $\alpha$ was constructed by parallel transport, so it had better be parallel — but we have to write it out carefully to see how the construction interacts with the definition of $\nabla$. Fix $y \in M$ and a tangent vector $v \in T_y M$, and choose any smooth curve $\eta : (-\varepsilon, \varepsilon) \to M$ with $\eta(0) = y$ and $\dot\eta(0) = v$. The covariant derivative $\nabla_v \alpha$ at $y$ is, by definition, the rate at which $\alpha$ deviates from being parallel along $\eta$, measured by pulling back via parallel transport. Concretely: \begin{align*} (\nabla_v \alpha)(y) = \frac{d}{ds}\bigg|_{s=0} (P_{\eta|_{[0,s]}})^{-1} \alpha_{\eta(s)}, \end{align*} where $(P_{\eta|_{[0,s]}})^{-1} : \mathcal{T}_{\eta(s)}^{(p,q)} \to \mathcal{T}_y^{(p,q)}$ pulls the value of $\alpha$ at $\eta(s)$ back to the fixed tensor space at $y$ so we can take a derivative. To evaluate this, we first need a formula for $\alpha_{\eta(s)}$ that is convenient near $s = 0$. Fix a path $\sigma_0$ from $x$ to $y$ used to compute $\alpha_y = P_{\sigma_0}\alpha_0$. For each $s$, let $\sigma_s := \sigma_0 * \eta|_{[0, s]}$ be the path from $x$ to $\eta(s)$ obtained by going $x \to y$ along $\sigma_0$ and then $y \to \eta(s)$ along the curve segment. By the well-definedness in Step 3, $\alpha$ at $\eta(s)$ may be computed using this $\sigma_s$, and by functoriality of parallel transport under concatenation, \begin{align*} \alpha_{\eta(s)} = P_{\sigma_s} \alpha_0 = P_{\eta|_{[0,s]}} \circ P_{\sigma_0} \alpha_0 = P_{\eta|_{[0,s]}} \alpha_y. \end{align*} This is the key identity: along the curve $\eta$, the values of $\alpha$ are exactly the parallel transport of $\alpha_y$. So pulling back via $(P_{\eta|_{[0,s]}})^{-1}$ undoes the transport: \begin{align*} (P_{\eta|_{[0,s]}})^{-1} \alpha_{\eta(s)} = (P_{\eta|_{[0,s]}})^{-1} P_{\eta|_{[0,s]}} \alpha_y = \alpha_y, \end{align*} which is independent of $s$. Differentiating in $s$ at $s = 0$ gives \begin{align*} (\nabla_v \alpha)(y) = \lim_{s \to 0} \frac{1}{s} \big[ (P_{\eta|_{[0,s]}})^{-1} \alpha_{\eta(s)} - \alpha_y \big] = \lim_{s \to 0} \frac{1}{s} \big[ \alpha_y - \alpha_y \big] = 0. \end{align*} Since $v$ and $y$ were arbitrary, $\nabla \alpha = 0$ on all of $M$. This is the cleanest possible verification of parallelism: $\alpha$ was *constructed* by parallel transport, so it is parallel. The construction is self-consistent precisely because of the well-definedness in Step 3, which is what required Hol-invariance of $\alpha_0$. The factorisation $P_{\sigma_s} = P_{\eta|_{[0,s]}} \circ P_{\sigma_0}$ uses the same functoriality of parallel transport as Step 3, and the first factor is what $\nabla_v$ measures — the local rate of change along $\eta$ near $s = 0$.[/guided]

[step:Combine the two directions to conclude the equivalence]The forward direction (Step 2) shows that condition 1 implies condition 2 with $\alpha_0 := \alpha_x$. The reverse direction (Steps 3–5) shows that condition 2 implies condition 1 with $\alpha$ defined by parallel transport from $x$. The two constructions are inverse to each other: starting from a parallel field $\alpha$, evaluating at $x$, then reconstructing yields the same field (by Step 5, the reconstructed field is parallel and has the right value at $x$, and parallel fields are determined by their value at any point on a connected manifold). Conversely, starting from $\alpha_0$, constructing $\alpha$, and re-evaluating at $x$ gives back $\alpha_0$ by Step 3. Hence conditions 1 and 2 are equivalent, and the bijection is \begin{align*} \{\alpha \in \Gamma(\mathcal{T}^{(p,q)} M) : \nabla\alpha = 0\} &\longleftrightarrow \{\alpha_0 \in \mathcal{T}_x^{(p,q)} : \alpha_0 \text{ is Hol-invariant}\}, \\ \alpha &\mapsto \alpha_x, \\ \alpha_y := P_\sigma(\alpha_0) &\mapsfrom \alpha_0. \end{align*} This is the assertion of the theorem.[/step]

[guided]We assemble the two directions into a single equivalence. Step 2 produced the forward map: from a parallel tensor field $\alpha$ (condition 1) we extract $\alpha_0 := \alpha_x$, which is Hol-invariant (condition 2). Steps 3–5 produced the reverse map: from a Hol-invariant $\alpha_0$ (condition 2) we constructed $\alpha$ by parallel transport from $x$, and verified it is well-defined, smooth, and parallel — so it satisfies condition 1. These two constructions are inverse to each other, which is what establishes the equivalence as a canonical bijection rather than just a logical biconditional. Going condition-1 $\to$ condition-2 $\to$ condition-1: given a parallel field $\alpha$, we extract $\alpha_x$ and reconstruct the parallel field by parallel-transporting $\alpha_x$ from $x$. The reconstructed field agrees with the original because parallel fields on a connected manifold are determined by their value at any single point — a parallel field along any curve is the parallel transport of its initial value, by the uniqueness of solutions to the parallel-transport ODE. Going condition-2 $\to$ condition-1 $\to$ condition-2: given a Hol-invariant $\alpha_0$, we build $\alpha$ and re-evaluate at $x$, obtaining $\alpha_x = \alpha_0$ — this is exactly what we verified at the end of Step 3 (parallel transport along the constant loop is the identity). Hence we have the canonical bijection \begin{align*} \{\alpha \in \Gamma(\mathcal{T}^{(p,q)} M) : \nabla\alpha = 0\} &\longleftrightarrow \{\alpha_0 \in \mathcal{T}_x^{(p,q)} : \alpha_0 \text{ is Hol-invariant}\}, \\ \alpha &\mapsto \alpha_x, \\ \alpha_y := P_\sigma(\alpha_0) &\mapsfrom \alpha_0, \end{align*} which is the assertion of the theorem. The bijection encodes a fundamental rigidity principle: a parallel tensor field is determined by *one* value on a connected manifold, and the only constraint on this value is invariance under the holonomy group. So the space of parallel tensor fields of a given type is a vector space isomorphic to the space of Hol-invariants in the corresponding tensor space at $x$. This has immediate consequences. For instance: - A parallel volume form on $(M, g)$ exists iff some $\mathrm{Hol}(M)$-invariant top form exists, which (since $\mathrm{Hol}(M) \subseteq \mathrm{O}(n)$) reduces to: the determinant character $\det : \mathrm{O}(n) \to \{\pm 1\}$ is trivial on $\mathrm{Hol}(M)$, i.e., $\mathrm{Hol}(M) \subseteq \mathrm{SO}(n)$. This recovers the equivalence "orientability via parallel volume form $\iff$ holonomy $\subseteq \mathrm{SO}(n)$". - An almost complex structure $J$ parallel under $\nabla$ exists iff there is an $\mathrm{Hol}(M)$-invariant complex structure on $T_x M$, i.e., $\mathrm{Hol}(M) \subseteq \mathrm{U}(n/2)$. - Special holonomy groups (Berger's list: $\mathrm{U}(m), \mathrm{SU}(m), \mathrm{Sp}(k), \mathrm{Sp}(k)\mathrm{Sp}(1), G_2, \mathrm{Spin}(7)$) correspond exactly to extra parallel tensor fields (Kähler form, complex volume, hyperkähler triple, etc.) that are pinned down by representation-theoretic invariants. This principle reduces the analytic question "does $(M, g)$ carry a parallel tensor field of a certain type?" to the algebraic question "does the holonomy representation of $\mathrm{Hol}_x(M)$ on $\mathcal{T}_x^{(p,q)}$ have a non-trivial fixed subspace?". It is the bridge between Riemannian geometry and the representation theory of orthogonal Lie groups — the foundation of the entire holonomy classification programme.[/guided]