Index form in constant curvature

For every field $V$ along $\gamma$ and every $t$ one has $g(V_{\perp },T)=0$ and $R(V,T)T=K{V}_{\perp }$. [F2, F4, given] By definition of $V_{\perp }$ and bilinearity of $g$, $g(V_{\perp },T)=g(V,T)-g(V,T)g(T,T)=0$ because $\gamma$ has unit speed, $g(T,T)=1$. Then [F2] gives $R(V,T)T=K(g(T,T)V-g(V,T)T)=K(V-g(V,T)T)=K{V}_{\perp }$.

In the setting of (b), the field $J(t)=s_K(t)E(t)$ is a nonzero Jacobi field with $J(0)=J(L)=0$. [F6, F7, F10, F11] The field $E$ is parallel and normal by [F6], so [F7] identifies $s_{K}E$ as a normal Jacobi field with $J(0)=0$. Since $s_K(0)=\sin 0\mathrm{/}\sqrt{K}=0$ and $s_K(L)=\sin (\pi )\mathrm{/}\sqrt{K}=0$ by [F11] and [F10], both endpoint values vanish; and $s_K(t)>0$ for $0<t<L$ by [F10] while $\mathrm{\mid }E(t)\mathrm{\mid }=1$, so $J$ is not the zero field.

Let $\phi (t)=\sin (\pi t\mathrm{/}L)$ for $0\le t\le L$. Then $\phi$ is smooth with $\phi (0)=\phi (L)=0$ and ${\phi }^{\mathrm{\prime }\mathrm{\prime }}(t)=-(\pi \mathrm{/}L{)}^2\phi (t)$. [F10, F11] The chain rule [F11] gives ${\phi }^{\mathrm{\prime }}(t)=(\pi \mathrm{/}L)\cos (\pi t\mathrm{/}L)$ and ${\phi }^{\mathrm{\prime }\mathrm{\prime }}(t)=-(\pi \mathrm{/}L{)}^2\sin (\pi t\mathrm{/}L)$. The endpoints use $\sin 0=0$ and $\sin \pi =0$.

For all continuous piecewise $C^1$ fields $V,W$ along $\gamma$, the index form is $I_{\gamma }(V,W)={\int }_a^b(g(D_{t}V,D_{t}W)-Kg(V_{\perp },W_{\perp }))dt$, and on the diagonal $I_{\gamma }(V,V)={\int }_a^b(g(D_{t}V,D_{t}V)-K\mathrm{\mid }{V}_{\perp }{\mathrm{\mid }}^2)dt$. [F3, F4, step 1.1] By step 1.1, $R(V,T)T=K{V}_{\perp }$ at every $t$; moreover $g(V_{\perp },W)=g(V_{\perp },W_{\perp })$ because $W-W_{\perp }=g(W,T)T$ is parallel to $T$ and $g(V_{\perp },T)=0$. Substituting into the defining sum of [F3] and using the diagonal identification $g(V_{\perp },V_{\perp })=\mathrm{\mid }{V}_{\perp }{\mathrm{\mid }}^2$ of [F4] gives both displayed identities, in particular on ${\mathcal{X}}_0(\gamma )$.

$I_{\gamma }(J,W)=0$ for every $W\in {\mathcal{X}}_0(\gamma )$. [F5, F8, F9, step 1.2] Both $J$ and $W$ are continuous on $[0,L]$, with $J$ smooth and $W$ piecewise $C^1$, so [F9] applies with no derivative jumps: the boundary term $[g(D_{t}J,W){]}_0^L$ vanishes because $W(0)=W(L)=0$, and the integral term vanishes pointwise because $D_t^{2}J+R(J,T)T=0$ by [F8] and step 1.2. Since $J$ is a nonzero element of the subspace ${\mathcal{X}}_0(\gamma )$, the fixed-endpoint index form on $[0,L]$ is degenerate.

In the setting of (c), $W(t)=\phi (t)E(t)$ lies in ${\mathcal{X}}_0(\gamma )$ and $I_{\gamma }(W,W)=((\pi \mathrm{/}L{)}^2-K){\int }_0^L{\phi }^{2}dt$. [F2, F5, F6, F9, F12, step 1.3] The field $W$ is smooth because $\phi$ and $E$ are, and $W(0)=W(L)=0$ by step 1.3, so $W\in {\mathcal{X}}_0(\gamma )$. Since $D_{t}E=0$ by [F6], one has $D_{t}W={\phi }^{\mathrm{\prime }}E$ and $D_t^{2}W={\phi }^{\mathrm{\prime }\mathrm{\prime }}E$; and $R(W,T)T=K(g(T,T)W-g(W,T)T)=K\phi E$ by [F2], because $g(E,T)=0$ and $g(T,T)=1$. So $D_t^{2}W+R(W,T)T=({\phi }^{\mathrm{\prime }\mathrm{\prime }}+K\phi )E$, and the pairing with $W$ is $({\phi }^{\mathrm{\prime }\mathrm{\prime }}+K\phi )\phi$. Applying [F9] with $V=W$ and no jumps, the boundary term is $[g(D_{t}W,W){]}_0^L=[{\phi }^{\mathrm{\prime }}\phi \mathrm{\mid }E{\mathrm{\mid }}^2{]}_0^L=0$ because $\phi (0)=\phi (L)=0$, hence $I_{\gamma }(W,W)=-{\int }_0^L({\phi }^{\mathrm{\prime }\mathrm{\prime }}+K\phi )\phi dt$. Step 1.3 gives ${\phi }^{\mathrm{\prime }\mathrm{\prime }}+K\phi =(K-(\pi \mathrm{/}L{)}^2)\phi$, so the last integral equals $(K-(\pi \mathrm{/}L{)}^2){\int }_0^L{\phi }^{2}dt$ by [F12], and its negative is the displayed value.

In the setting of (c), ${\int }_0^L{\phi }^{2}dt>0$. [F10, F13, F14, step 1.3] The function ${\phi }^2$ is continuous and nonnegative on $[0,L]$. At the midpoint $t=L\mathrm{/}2$ one has $\phi (L\mathrm{/}2)=\sin (\pi \mathrm{/}2)>0$ by [F10], so ${\phi }^2$ is not identically zero; if its integral vanished, [F14] would force $\phi \equiv 0$, a contradiction. Therefore the integral is nonzero, and it is $\ge 0$ by [F13]; hence it is strictly positive.

If $L>\pi \mathrm{/}\sqrt{K}$ then $I_{\gamma }(W,W)<0$. [step 2.3, step 2.4, algebra] By step 2.3, $I_{\gamma }(W,W)=((\pi \mathrm{/}L{)}^2-K){\int }_0^L{\phi }^{2}dt$. The assumption $L>\pi \mathrm{/}\sqrt{K}$ makes $(\pi \mathrm{/}L{)}^2-K<0$, and step 2.4 makes the second factor positive, so the product is negative. Therefore the fixed-endpoint index form on a unit-speed segment of length $L>\pi \mathrm{/}\sqrt{K}$ is not positive semidefinite.

Boundary, degeneracy and choice audit. [A1, F1, F3, F5, F9, step 1.1, step 2.1, step 2.2, step 3.1] The interval $[a,b]$ of (a) is nondegenerate and included endpoints carry the one-sided derivatives of [F5]. In (b) and (c) the length $L>0$ is fixed and $0$ is an included endpoint; at the critical length $L=\pi \mathrm{/}\sqrt{K}$ the case (c) is excluded and case (b) shows only degeneracy, not negativity. The formulas are vacuous in the empty manifold and in dimension zero, where no unit-speed geodesic exists; in dimension one every field is parallel to $T$, so $V_{\perp }=0$, part (a) reduces to $I_{\gamma }(V,V)=\int \mathrm{\mid }{D}_{t}V{\mathrm{\mid }}^{2}dt$, and parts (b) and (c) are vacuous because no normal direction exists. For $K=0$ part (a) gives the flat formula $I_{\gamma }(V,W)=\int g(D_{t}V,D_{t}W)dt$, while (b) and (c) require $K>0$. The zero field has zero index form, and the nonzero field $J$ of step 2.2 is the explicit degeneracy witness. Assumption [A1] is inherited from the index-form and constant-curvature suppliers; the two test fields are given by explicit formulas and no selection is made. The example asserts identities and one-way implications and claims no equivalence. $\mathrm{\square }$