# Tensor well-defined on a manifold, basic concepts.

binbagsss
I'm trying to understand what exactly it means by some tensor field to be 'well-defined' on a manifold. I'm looking at some informal definition of a manifold taken to be composed of open sets ##U_{i}##, and each patch has different coordinates.

The text I'm looking at then talks about how in order for a tensor field to be defined globally there are certain transition laws that must be obeyed in intersecting regions of ##U_{i}##

From this, my interpretation of well-defined is that you have in variance in certain patches, and so because each patch has its own coordinates, you have in variance in any coordinates. So, a tensor means you have invariance with respect to a change in coordinate system?Are these thoughts correct?

Is this literal agreement? I.e- the value of a scalar? the components of a matrix (representing a tensor as a matrix)?

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Staff Emeritus
Gold Member
To say that something is "well defined" typically means that whatever you have previously said about it is sufficient to actually define it. So you may need to tell us that we what said previously.

I will make a guess based on what you said. I would define a vector field on a smooth manifold ##M## as a function that associates a tangent vector at p with each ##p\in M##. It sounds like their version of this is a function that associates an n-tuple of real numbers with each pair ##(p,x)## such that ##x:U\to\mathbb R^n## is a coordinate system such that ##p\in U\subseteq M##. This type of function indirectly defines a tensor, if and only if for all ##p\in M## and all coordinate systems ##x:U\to\mathbb R^n## and ##y:V\to\mathbb R^n## such that ##p\in U\cap V##, the relationship between the n-tuple associated with ##(p,x)## and the n-tuple associated with ##(p,y)## is given by the tensor transformation law.

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Staff Emeritus