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Matthias Ludewig
Matthias Ludewig
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This has nothing to do with Lie groups, I believe. Let $M$ be a Riemannian manifold. The Bochner formula on $1$-forms states that $$\nabla^* \nabla \omega = (d \delta + \delta d)\omega - \mathrm{Ric}\,\omega.$$ Hence we have for any compactly supported function $f$ (writing round brackets for the $L^2$ scalar product) $$\|Hf\|_{L^2}^2 = \|\nabla d f\|_{L^2}^2 = (\nabla^* \nabla d f, df) = ((d \delta + \delta d + \mathrm{Ric})d f, df ) = (d\delta d f, d f) - (\mathrm{Ric}\, df, df)$$$$\|Hf\|_{L^2}^2 = \|\nabla d f\|_{L^2}^2 = (\nabla^* \nabla d f, df) = ((d \delta + \delta d - \mathrm{Ric})d f, df ) = (d\delta d f, d f) - (\mathrm{Ric}\, df, df)$$ Using that on functions, $\Delta f = \delta d f$, this is equal to $$(d \Delta f, df) - (\mathrm{Ric}\, df, df) = \|\Delta f\|_{L^2}^2 - (\mathrm{Ric}\, df, df)$$ This is a standard formula of Riemannian geometry.

This has nothing to do with Lie groups, I believe. Let $M$ be a Riemannian manifold. The Bochner formula on $1$-forms states that $$\nabla^* \nabla \omega = (d \delta + \delta d)\omega - \mathrm{Ric}\,\omega.$$ Hence we have for any compactly supported function $f$ (writing round brackets for the $L^2$ scalar product) $$\|Hf\|_{L^2}^2 = \|\nabla d f\|_{L^2}^2 = (\nabla^* \nabla d f, df) = ((d \delta + \delta d + \mathrm{Ric})d f, df ) = (d\delta d f, d f) - (\mathrm{Ric}\, df, df)$$ Using that on functions, $\Delta f = \delta d f$, this is equal to $$(d \Delta f, df) - (\mathrm{Ric}\, df, df) = \|\Delta f\|_{L^2}^2 - (\mathrm{Ric}\, df, df)$$ This is a standard formula of Riemannian geometry.

This has nothing to do with Lie groups, I believe. Let $M$ be a Riemannian manifold. The Bochner formula on $1$-forms states that $$\nabla^* \nabla \omega = (d \delta + \delta d)\omega - \mathrm{Ric}\,\omega.$$ Hence we have for any compactly supported function $f$ (writing round brackets for the $L^2$ scalar product) $$\|Hf\|_{L^2}^2 = \|\nabla d f\|_{L^2}^2 = (\nabla^* \nabla d f, df) = ((d \delta + \delta d - \mathrm{Ric})d f, df ) = (d\delta d f, d f) - (\mathrm{Ric}\, df, df)$$ Using that on functions, $\Delta f = \delta d f$, this is equal to $$(d \Delta f, df) - (\mathrm{Ric}\, df, df) = \|\Delta f\|_{L^2}^2 - (\mathrm{Ric}\, df, df)$$ This is a standard formula of Riemannian geometry.

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Matthias Ludewig
Matthias Ludewig
  • 9.9k
  • 1
  • 30
  • 71

This has nothing to do with Lie groups, I believe. Let $M$ be a Riemannian manifold. The Bochner formula on $1$-forms states that $$\nabla^* \nabla \omega = (d \delta + \delta d)\omega - \mathrm{Ric}\,\omega.$$ Hence we have for any compactly supported function $f$ (writing round brackets for the $L^2$ scalar product) $$\|Hf\|_{L^2}^2 = \|\nabla d f\|_{L^2}^2 = (\nabla^* \nabla d f, df) = ((d \delta + \delta d + \mathrm{Ric})d f, df ) = (d\delta d f, d f) + (\mathrm{Ric}\, df, df)$$$$\|Hf\|_{L^2}^2 = \|\nabla d f\|_{L^2}^2 = (\nabla^* \nabla d f, df) = ((d \delta + \delta d + \mathrm{Ric})d f, df ) = (d\delta d f, d f) - (\mathrm{Ric}\, df, df)$$ Using that on functions, $\Delta f = \delta d f$, this is equal to $$(d \Delta, df) + (\mathrm{Ric}\, df, df) = \|\Delta f\|_{L^2}^2 + (\mathrm{Ric}\, df, df)$$$$(d \Delta f, df) - (\mathrm{Ric}\, df, df) = \|\Delta f\|_{L^2}^2 - (\mathrm{Ric}\, df, df)$$ This is a standard formula of Riemannian geometry.

This has nothing to do with Lie groups, I believe. Let $M$ be a Riemannian manifold. The Bochner formula on $1$-forms states that $$\nabla^* \nabla \omega = (d \delta + \delta d)\omega - \mathrm{Ric}\,\omega.$$ Hence we have for any compactly supported function $f$ (writing round brackets for the $L^2$ scalar product) $$\|Hf\|_{L^2}^2 = \|\nabla d f\|_{L^2}^2 = (\nabla^* \nabla d f, df) = ((d \delta + \delta d + \mathrm{Ric})d f, df ) = (d\delta d f, d f) + (\mathrm{Ric}\, df, df)$$ Using that on functions, $\Delta f = \delta d f$, this is equal to $$(d \Delta, df) + (\mathrm{Ric}\, df, df) = \|\Delta f\|_{L^2}^2 + (\mathrm{Ric}\, df, df)$$ This is a standard formula of Riemannian geometry.

This has nothing to do with Lie groups, I believe. Let $M$ be a Riemannian manifold. The Bochner formula on $1$-forms states that $$\nabla^* \nabla \omega = (d \delta + \delta d)\omega - \mathrm{Ric}\,\omega.$$ Hence we have for any compactly supported function $f$ (writing round brackets for the $L^2$ scalar product) $$\|Hf\|_{L^2}^2 = \|\nabla d f\|_{L^2}^2 = (\nabla^* \nabla d f, df) = ((d \delta + \delta d + \mathrm{Ric})d f, df ) = (d\delta d f, d f) - (\mathrm{Ric}\, df, df)$$ Using that on functions, $\Delta f = \delta d f$, this is equal to $$(d \Delta f, df) - (\mathrm{Ric}\, df, df) = \|\Delta f\|_{L^2}^2 - (\mathrm{Ric}\, df, df)$$ This is a standard formula of Riemannian geometry.

Source Link
Matthias Ludewig
Matthias Ludewig
  • 9.9k
  • 1
  • 30
  • 71

This has nothing to do with Lie groups, I believe. Let $M$ be a Riemannian manifold. The Bochner formula on $1$-forms states that $$\nabla^* \nabla \omega = (d \delta + \delta d)\omega - \mathrm{Ric}\,\omega.$$ Hence we have for any compactly supported function $f$ (writing round brackets for the $L^2$ scalar product) $$\|Hf\|_{L^2}^2 = \|\nabla d f\|_{L^2}^2 = (\nabla^* \nabla d f, df) = ((d \delta + \delta d + \mathrm{Ric})d f, df ) = (d\delta d f, d f) + (\mathrm{Ric}\, df, df)$$ Using that on functions, $\Delta f = \delta d f$, this is equal to $$(d \Delta, df) + (\mathrm{Ric}\, df, df) = \|\Delta f\|_{L^2}^2 + (\mathrm{Ric}\, df, df)$$ This is a standard formula of Riemannian geometry.