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The Laplacian matrix is the representation of a graph in matrix form.
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Green kernel vs fundamental solution
Let $L$ being the Laplacian for a given Lie group $G$. …
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Laplacian on two Lie groups have the same Lie algebra
Then, the Casimir-Laplacian of $G$ is given by
$$\Delta_{G}= \sum_{i=1}^{n} X_i^2.$$
Now, if $G_1$ is a Lie group that has the same Lie algebra of another Lie group $G_2$. …
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How to prove that the Laplace oparator of the Heisenberg group is sub-elliptic?
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For $z=x+ i y \in \mathbb C$ and $t\in \mathbb R$, the Laplacian operator on the Heisenberg group $H^3$ is given by
$$ \Delta= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} … + (x^2+y^2 ) \frac{\partial^2}{\partial t^2} + 2(x\frac{\partial}{\partial y} -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t} .$$
I want to show that the Laplacian operator $ \Delta$ is …
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Difference between the Laplacian and the sub-Laplacian of a Lie group
Given a Lie group $G$, what is the difference between the Laplacian $\Delta$ and the sub-Laplacian $\Delta_{sub}$ of $G$. … For example, what I know, for $G$ being the Heisenberg group $H^3= \mathbb C \times \mathbb R$, the difference between the Laplacian $\Delta$ and the sub-Laplacian $\Delta_{sub}$ of $H^3$ is the standard …
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Why the sub-Laplacian $\Delta_{sub}$ on the Heisenberg group $H^3 $ is sub-elliptic but not ...
I want to know why the sub-Laplacian $\Delta_{sub}= X^2 + Y^2$ on the Heisenberg group $H^3 = \mathbb C \times \mathbb R$ is sub-elliptic but not elliptic, where $X$ and $Y$ are the left-invariant vector … And what has been the same for the sub-Laplacian on the quaternionic Heisenberg group $H^7 = \mathbb H \times \mathbb R^3$, where $\mathbb H$ is the space of the quaternion.
Thank you in advance …