I am currently reading about tensor fields on manifolds, and I came across two comments that sound contradictory to me.

The first comment is made in the book by James Munkres "Analysis on Manifolds", page 249:

Any tensor field on M [M a manifold and a subset of $\mathbb{R}^n$] can be extended to a tensor field defined on an open set of $\mathbb{R}^n$ containing $M$.

The second comment is made in the book by John Lee "Riemannian Manifolds, An Introduction to Curvature", page 56:

Not every vector field along a curve [in a manifold M] need be extendible [to a neighborhood of the curve in M].

As an example, Lee mentions the case of a vector field of an intersecting curve.

Now - the notion of a tensor is quite new to me so I might go wrong here but I thought a vector field is a contravariant tensor field of rank 1. But if that is the case then the above comments contradict each other... a clarification of where I went wrong in my reasoning would be a great help, many thanks!

  • 2
    $\begingroup$ A curve is not always an embedded manifold, for example, if it is self-intersecting. Even when it is an embedded manifold, extension might not work: consider the open interval $(0, 1)$ embedded as a broken circle in $\mathbb{R}^2$. $\endgroup$ – Zhen Lin Apr 6 '12 at 23:49
  • $\begingroup$ So wouldn't it seem that the broken circle also a counterexample to the first statement by Munkres? $\endgroup$ – Nick Alger Apr 7 '12 at 0:21
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    $\begingroup$ Actually, any tensor field on the broken circle can be extended to an open subset of $\mathbb R^n$. For example, if $M\subset\mathbb R^2$ is the unit circle minus $\{(1,0)\}$, then any tensor field on it can be extended to the plane minus the nonnegative $x$-axis. As Zhen Lin said, the difference arises because in my Riemannian manifolds book, a "curve" need not be a submanifold because it isn't necessarily injective. $\endgroup$ – Jack Lee Apr 10 '12 at 17:25

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