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According to Guillemin and Pollack's Differential Topology:

A compact interval $X = [0,1]$ inherite standard orientation from $\mathbb{R}$.

To my understanding, "inherit from $\mathbb{R}$" means a linear transformation between $X$ and the standard basis of $\mathbb{R}$ whose determinant is positive, right? Then, to start with, what is the standard basis of $\mathbb{R}$? Is it just $1$?

Thanks.

Seirios
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1LiterTears
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1 Answers1

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Yes, the standard basis of $\mathbb R$ is just $1$ - or more generally the standard basis of $\mathbb R^n$ consists of the vectors $(1,0\ldots0), (0,1,0\ldots,0), \ldots,(0,\ldots,0,1)$ in this order.

Since $X$ is a subspace of $\mathbb R$, the tangent space of $X$ in any point $x\in X$ is a subspace of the tangent space of $\mathbb R$ in that same point $x$ (i.e. \mathbb R again), and for interior points they in fact coincide just by looking at the dimension and hence "everything" tangential (including orientation) is inerited by $X$ (except at its boundary).

Hagen von Eitzen
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  • OH, that's really helpful, thanks Hagen. So basically, the orintation $[0,1]$ inherited from $\mathbb{R}$ is just the one that the 1 boundary is positive, right? Since we transform with an identity map from $[0,1] \subset \mathbb{R}$ to $[0,1]$. The determinant of identity is positive, hence its orientation is inherited from $\mathbb{R}$ to $[0,1]$. – 1LiterTears Jul 19 '13 at 06:22