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If $F_\nabla$ is the curvature associated to a connection $\nabla$ on a 2-dimensional Riemannian manifold $M$ with metric $g$, then we can define ''the'' Riemann curvature by (https://en.wikipedia.org/wiki/Curvature_of_Riemannian_manifolds) by $$ R(X,Y,Z,W) = g(F_\nabla (X,Y)Z, W) $$ for vector fields $X,Y,Z,W$. This is related to a function $K: M\to \mathbb{R}$ by $$ K(x)\sigma(X,Y)\sigma(Z,W) = R(X,Y,Z,W)\mid_x $$ where $\sigma = e_1^{*}\wedge e_2^*$ is a globally-defined non-vanishing 2-form, and $e_1^*, e_2^*$ are dual to an orthonormal frame $e_1, e_2$. (Side note: is this function the same as the sectional curvature?)

What I would like to understand is whether there is a more direct way to express $K$ in terms of $g$ and $F_\nabla$: for example, a dual section $s^*$ is defined by $s^{*} = g(s,\cdot)$ - so can we say something like $$ K(x)\sigma = \langle\omega , e_1\wedge e_2\rangle\mid_x $$ where $\omega(s_1, s_2) := g(F_\nabla s_1, s_2)$ for vector fields $s_1, s_2$, and $\langle \cdot, \cdot\rangle$ in this instance is meant to indicate the natural pairing of a 2-form with a bivector.

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  • $\begingroup$ Is $\sigma$ meant to be the Riemannian area form? As written, it seems there are many different combinations $(K,\sigma)$ that could satisfy your relation with $R$. $\endgroup$ – Anthony Carapetis Feb 7 '18 at 4:48
  • $\begingroup$ @AnthonyCarapetis, not necessarily; $\sigma$ is just defined as above with no extra conditions, though I guess perhaps the Riemannian area form might make $K$ the sectional curvature. The core of my question is whether it's possible to manipulate duals and metrics in the manner I spoke of above to derive the last pointwise equality for some function $K$ satisfying the given condition. $\endgroup$ – An Coileanach Feb 7 '18 at 14:13
  • $\begingroup$ Is $e_1^*,e_2^*$ the dual frame of $e_1,e_2$ (in the sense of a dual basis), or is it $e_i^* = g(e_i, \cdot)$? These are the same (and $\sigma$ is $\pm$ the area form) if $e_i$ is orthonormal, but otherwise they differ. $\endgroup$ – Anthony Carapetis Feb 8 '18 at 11:14
  • $\begingroup$ Also, I assume the brackets in your last equation are meant to be the natural pairing of a 2-form with a bivector? i.e. it's just $K \sigma(X,Y) = g(F(X,Y)e_1,e_2)$? $\endgroup$ – Anthony Carapetis Feb 8 '18 at 11:21
  • $\begingroup$ @AnthonyCarapetis, $e_1^*, e_2^*$ is the dual frame of an orthonormal frame. You are also correct that by $\langle \omega, e_1\wedge e_2 \rangle$, I meant the natural pairing of a 2-form with a bivector. Apologies for the lack of clarity. I will amend my question accordingly. $\endgroup$ – An Coileanach Feb 8 '18 at 18:35
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By the definition of the wedge product we have $\sigma(e_1,e_2)=e_1^*(e_1)e_2^*(e_2)-e_2^*(e_1)e_1^*(e_2)=1$; so just plugging this frame in to the equation $K\, \sigma \otimes \sigma = R$ yields the explicit formula $$R(e_1,e_2,e_1,e_2)=K \sigma(e_1,e_2)\sigma(e_1,e_2) = K.$$ Since $e_i$ is an orthonormal frame, this is indeed the sectional curvature; and also $\sigma$ is the Riemannian area form up to sign. (As an aside, unless your surface is genus $1$ there is no global orthonormal frame; so the equation $\sigma = e_1^* \wedge e_2^*$ should be understood locally.)

If you think of $R$ as a symmetric bilinear form on bivectors then this can be written $$K = R(e_1 \wedge e_2, e_1 \wedge e_2) = R(\sigma^*, \sigma^*).$$

Here the the metric dual relating bivectors and two-forms can be simply defined by $v\wedge w \mapsto v^* \wedge w^*,$ or alternatively by $$(v \wedge w)^* = g_2(v \wedge w,\cdot)$$ where $$g_2(v\wedge w, x \wedge y) = g(v,x)g(w,y) - g(v,y)g(w,x)$$ is the inner product on bivectors induced by $g$.

Your proposed equation for $K\sigma$ is what we get by only plugging in the frame once instead of twice: $$R(e_1,e_2,X,Y) = K\sigma(e_1,e_2)\sigma(X,Y) = K\sigma(X,Y),$$ which in terms of bivectors becomes $$K \sigma = R(e_1\wedge e_2, \cdot).$$ Your $\omega$ is really the same thing as $R$, and the natural pairing is the same thing as the partial evaluation I have written.

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  • $\begingroup$ Thank you so much! Your very clear explanation has really helped my understanding. I only wish I could upvote this answer multiple times. Thanks for your patience in dealing with my poorly-phrased question. $\endgroup$ – An Coileanach Feb 9 '18 at 7:57
  • $\begingroup$ Glad I could help! :) $\endgroup$ – Anthony Carapetis Feb 9 '18 at 8:50

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