This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Orientation in Euclidean Space
Consider the standard basis \((e_1,e_2,e_3)\) in three-dimensional space. We have known since learning calculus that the order in which this basis is arranged matters. For example, the above basis satisfies \(e_1\times e_2=e_3\), but if the order were changed and arranged as \((e_2,e_1,e_3)\), then we would have \(e_2\times e_1=-e_3\).
We do not need to restrict this observation to the standard basis. In general, for any orthonormal basis \((x_1,x_2,x_3)\) of \(\mathbb{R}^3\), we can tell whether this basis is arranged in the correct order or not according to whether the value of \(x_1\times x_2\) is \(x_3\) or \(-x_3\). Writing this as a formula, the value of
\[x_3\cdot(x_1\times x_2)\]being \(+1\) or \(-1\) determines the order. However, the above expression equals
\[x_3\cdot(x_1\times x_2)=\det[x_3\;x_1\;x_2]=\det[x_1\;x_2\;x_3]\]so we can determine the order for a general basis that is not necessarily orthonormal by reading whether the value of the above determinant is positive or negative.
Once we define the order of a basis via the determinant in the case of \(\mathbb{R}^3\), we can naturally tell in \(\mathbb{R}^m\) as well whether a basis \((x_1,\ldots, x_m)\) is arranged in the correct order or in the opposite order. Namely, we investigate the sign of the determinant
\[\det[x_1\;x_2\;\cdots\;x_m]\]Determinant and Orientation
Let \(V,W\) be \(n\)-dimensional \(\mathbb{R}\)-vector spaces and let \(L:V\rightarrow W\) be a linear map. Then, by the universal property of [Multilinear Algebra] §Tensor Algebras, ⁋Proposition 11, the following linear map
\[\bigwedge\nolimits^n(L):\bigwedge\nolimits^n(V)\rightarrow\bigwedge\nolimits^n(W)\]is well-defined. On the other hand, since both \(V\) and \(W\) are \(n\)-dimensional, \(\bigwedge\nolimits^n(V)\) and \(\bigwedge\nolimits^n(W)\) are both one-dimensional vector spaces, and therefore the above linear map is uniquely determined by where any nonzero vector is sent. In particular, if \(V=W\), then any nonzero vector in \(\bigwedge\nolimits^n(V)\) is always sent to a scalar multiple of itself, and this scalar equals the determinant of \(L\). From this perspective, \(\bigwedge\nolimits^n(L)\) is sometimes called the determinant map, and if \(E\) is an \(n\)-dimensional vector bundle defined over a manifold \(M\), then \(\bigwedge\nolimits^n(E)\) is called the determinant bundle of \(E\).
In particular, if \(E=T^\ast M\), we define it as follows.
Definition 1 Let \(M\) be an \(m\)-dimensional connected manifold. Then \(M\) is said to be orientable if \(\bigwedge\nolimits^m(M)\setminus\{0\}\) has two components, and choosing one of the two components is called an orientation of \(M\).
Proposition 2 Let \(M\) be an \(m\)-dimensional connected manifold. Then the following are all equivalent.
- \(M\) is orientable.
- There exists a suitable collection of coordinate systems covering \(M\) such that the Jacobian is always positive on their overlaps.
- There exists a non-vanishing \(m\)-form defined on \(M\).
Proof
Example 3
Any Lie group is orientable. This is because if we choose any basis \(\omega_1,\ldots,\omega_n\) in \(\Omega_\text{l.inv}^\ast(G)\) and consider their wedge \(\omega_1\wedge\cdots\wedge\omega_n\), this defines a nonvanishing \(n\)-form on \(G\).
References
[War] Frank W. Warner. Foundations of Differentiable Manifolds and Lie Groups, Graduate texts in mathematics, Springer, 2013
[Lee] John M. Lee. Introduction to Smooth Manifolds, Graduate texts in mathematics, Springer, 2012
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