On pages 118-119 of Spivak's Calculus on Manifolds he introduces the idea of consistent orientations by stating

It is often necessary to choose an orientation $\mu_x$ for each tangent space $M_x$ of a manifold $M \subset \mathbb R^n$. Such choices are called consistent provided that for every coordinate system $f:W\rightarrow M \subset \mathbb{R}^n$ and $a,b\in W$ the relation $$[f_*((e_1)_a),\ldots,f_*(e_k)_a)]=\mu_{f(a)}$$ holds if and only if $$[f_*((e_1)_b),\ldots,f_*(e_k)_b)]=\mu_{f(b)}.$$ Suppose orientations $\mu_x$ have been chosen consistently. If $f:W\rightarrow M \subset \mathbb{R}^n$ is a coordinate system such that $$[f_*((e_1)_b),\ldots,f_*((e_k)_b)]=\mu_{f(a)}$$ for one, and hence every $a\in W$, then $f$ is called orientation-preserving.

He then goes on to say

If $f$ and $g$ are orientation-preserving and $x=f(a)=g(b)$, then the relation $$[f_*((e_1)_a),\ldots,f_*((e_k)_a)]=\mu_x=[g_*((e_1)_b),\ldots,g_*((e_k)_b)]$$ implies that $$[(g^{-1}\circ f)_*((e_1)_a,\ldots,(e_k)_a)]=[(e_1)_b,\ldots,(e_k)_b],$$ so that $\det\,(g^{-1}\circ f)'>0$, an important fact to remember.

It is this last part that slightly confuses me. It seems intuitively clear, but a rigorous proof of this has eluded me. I have a few questions regarding the matter. Why is Spivak able to apply $(g^{-1})_*$ and keep the bases in the same orientation class? How do we know that $(g^{-1})_*$ applied to the vectors $f_*((e_i)_a)$ and $g_*((e_i)_b)$ for $i=1,\ldots,k$ doesn't impact the orientations in different ways (i.e. reverse one and preserve the other)? Also, if it is true that it doesn't influence the orientations differently, then does this same fact apply to $g_*$ and other coordinate systems for that matter? I believe that $(g^{-1})_*g_*=I_k$ (the $k\times k$ identity matrix, not sure if helpful). Lastly, I know we call $f,g$ orientation preserving, but does $[f_*((e_1)_a),\ldots,f_*((e_k)_a)]=\mu_{f(a)}$ imply $[f_*((v_1)_a),\ldots,f_*((v_k)_a)]=\mu_{f(a)}$ for any basis vectors $v_i$ with the same orientation as the standard basis on $\mathbb{R}^k$?

Note: The square brackets are used by Spivak to represent the equivalence class for the orientation of a given basis. The orientation to which a basis $\{v_1,\ldots,v_k\}$ belongs is denoted $$[v_1,\ldots,v_k]$$ and the other orientation is denoted $$-[v_1,\ldots,v_k].$$ If $v_1,\ldots,v_k$ and $w_1,\ldots,w_k$ are two bases and $A=(a_{ij})$ is defined by $w_i=\sum a_{ij}v_j$, then $v_1,\ldots,v_k$ and $w_1,\ldots,w_k$ are in the same orientation class if and only if $\det(A)>0$. Also, $e_1,\ldots,e_k$ denote the standard basis vectors in $\mathbb{R}^k$.

  • $\begingroup$ Your question is missing the definition of the "square bracket" notation: What is the definition of $[w_1,...,w_k]$, given an ordered basis $w_1,...,w_k$ of $\mathbb R^k$? Perhaps if you ponder that definition then the answer will be clear. If not, you should include that definition in your post. $\endgroup$
    – Lee Mosher
    Apr 21, 2021 at 12:41
  • $\begingroup$ The square bracket is just an equivalence class for the basis orientation used in the book mentioned in the post. $\endgroup$
    – user866625
    Apr 21, 2021 at 13:08
  • $\begingroup$ Yes, but that is just notation. How is the equivalence relation itself defined? Given, say, two ordered bases $v_1,...,v_k$ and $w_1,...,w_k$ of $\mathbb R^k$, what is the definition of the statement "the ordered basis $v_1,...,v_k$ is equivalent to the ordered basis $w_1,...,w_k$" ? $\endgroup$
    – Lee Mosher
    Apr 21, 2021 at 13:24
  • $\begingroup$ If $v_1,\ldots,v_k$ and $w_1,\ldots,w_k$ are two bases and $A=(a_{ij})$ is defined by $w_i=\sum a_{ij}v_j$, then $v_1,\ldots,v_k$ and $w_1,\ldots,w_k$ are in the same orientation class if and only if $\det(A)>0$. $\endgroup$
    – user866625
    Apr 21, 2021 at 14:05
  • $\begingroup$ Please do include this definition, as other (equivalent) definitions are available, and a possible answer heavily depends on this. However, I'm confused by two things. (a) What are $e_1,\ldots,e_k$? (b) $g^{-1}\circ f$ seems to be a map from the manifold into the manifold, and I understand that $(g^{-1}\circ f)'$ is supposed to be the differential. But what would be the determinant in this case? $\endgroup$ Apr 21, 2021 at 15:16

1 Answer 1


The general linear algebra fact that you need here is this:

Theorem: If $T : V_1 \to V_2$ is a linear isomorphism between two $k$-dimensional vector spaces, if $v_1,\ldots,v_k$ and $w_1,\ldots,w_k$ are two bases of $V_1$, and if $[v_1,\ldots,v_k]=[w_1,\ldots,w_k]$, then $[T(v_1),...,T(v_k)]=[T(w_1),\ldots,T(w_k)]$.

To prove this, let $A$ be the change of basis matrix from the $v$'s to the $w$'s, i.e. $w_i = \sum_j a_{ij} v_j$. From the assumption that $[v_1,\ldots,v_k]=[w_1,\ldots,w_k]$ it follows that $\det(A)>0$. Using the properties of linear transformations it follows that $$T(w_i) = T\left(\sum_j a_{ij} v_j\right) = \sum_j a_{ij} T(v_j) $$ Therefore $A$ is also the change of basis matrix from the $T(v)$'s to the $T(w)$'s. Since $\det(A)>0$, it follows that $[T(v_1),...,T(v_k)]=[T(w_1),\ldots,T(w_k)]$.

One can apply this in your setting using $T = (g^{-1})_*$, $V_1 = T_x M$, and $V_2 = T_b W = T_b \mathbb R^k = \mathbb R^k$, to derive the implication in your second quotation from Spivak.

  • $\begingroup$ Thank you! I was using notation from Spivak, but I know his notation is not always conventional. Does it matter that the $w_i$'s and $v_j$'s in your post are in a $k$-dimensional subspace (the tangent space to the $k$-manifold) of an $n$-dimensional vector space ($\mathbb{R}^n$)? I wasn't sure if we could construct a change of basis matrix since it would be a $k\times k$ matrix, but the vectors we would be multiplying by would be living in $\mathbb{R]^n$ technically. $\endgroup$
    – user866625
    Apr 21, 2021 at 23:37
  • $\begingroup$ You can define a change of basis matrix between any two bases in any (finite dimensional) vector space. It doesn't matter whether the vector space is $\mathbb R^k$ or something more abstract such as a tangent space at some point of a $k$-dimensional manifold. All you're doing to define it is to use the properties of a basis: $w_i$ can be expressed uniquely as a linear combination of the $v_j$'s, so those coefficients $a_{ij}$ are well-defined. $\endgroup$
    – Lee Mosher
    Apr 22, 2021 at 2:06
  • $\begingroup$ That makes sense I suppose. Thanks again for the answer and timely responses. $\endgroup$
    – user866625
    Apr 22, 2021 at 2:17

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