Onyx
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- How do I find the metric tensor on ##S^1## x ##S^2##?
How do I find the metric tensor on ##S^1## x ##S^2##?
The metric tensor on the manifold ##S^1 \times S^2## can be derived from the standard metrics of the circle and the sphere. The product metric for this manifold is expressed as ##d\theta^2 + d\psi^2 + \sin^2\theta d\phi^2##. This formulation is analogous to the metrics in the plane ##\mathbb{R}^2##, which can be represented as ##dx^2 + dy^2## or in polar coordinates as ##dr^2 + r^2 d\theta^2##. Understanding these relationships is crucial for working with product manifolds in differential geometry.
PREREQUISITESMathematicians, physicists, and students of differential geometry who are interested in understanding the properties and applications of metric tensors on complex manifolds.
What is the metric in the plane ##\mathbb R^2##?Onyx said:How do I take the product metric of the circle and sphere metrics?
Well, the plane ##\mathbb R^2## is the product ##\mathbb R \times \mathbb R## and the ##dx^2## and ##dy^2## are the metrics on each factor.Onyx said:##dx^2+dy^2## or ##dr^2+r^2d\theta^2##.
Well then I suppose for ##S^1 x S^2## it would be ##d\theta^2+d\psi^2+sin^2\theta d\phi^2##.martinbn said:Well, the plane ##\mathbb R^2## is the product ##\mathbb R \times \mathbb R## and the ##dx^2## and ##dy^2## are the metrics on each factor.