In Guillemin and Pollack, Differential Topology Page 166,

The automatic appearance of the compensating factor $\det (df)$ is a mechanical consequence of the anticommuntative behavior of $1$-forms: $$dx_i \wedge dx_j = - dx_j \wedge dx_i.$$

However, to my understanding, it is the absolute value of $\det(df)$ that compensates the anticommutativity. Could someone clarify this for me?

Thank you.

  • $\begingroup$ anticommutativity of what? The absolute value of the determinant of the Jacobian matrix (commonly called the Jacobian) arises in connection with the change of variables theorem for multivariate calculus. It is there because the area, volume etc... integrals are not oriented hence any sign generated by a change of orientation must be squashed. That's my intuition. The careful proof takes a few pages... $\endgroup$ – James S. Cook Aug 7 '13 at 1:45
  • $\begingroup$ Hello @JamesS.Cook, I see how $A^* \phi_1 \wedge \cdots \wedge A^* \phi = (\det A)\phi_1 \wedge \cdots \wedge \phi_k$. However, for this part, I just think the anticommutivity deals with the sign, not the determinant. $\endgroup$ – 1LiterTears Aug 7 '13 at 2:26
  • $\begingroup$ I wonder if math.stackexchange.com/questions/459658/… would help bring light to that formula. The question is a bit different, but the determinant arises for more or less the same reason I think...I'm not entirely sure about your comment here. $\endgroup$ – James S. Cook Aug 7 '13 at 2:47
  • $\begingroup$ How inspring @JamesS.Cook! Thank you so much for your reference. All your posts are so excellent and help me a lot! A quick question though - what do you mean exactly by "James annti-symmetric symbol"? Thank you! $\endgroup$ – 1LiterTears Aug 7 '13 at 3:05
  • $\begingroup$ the antisymmetric symbol is defined to be $1$ if the indices are in order from 1 to n; $\epsilon_{12\dots n}=1$ then all other nonzero values can be obtained by multiplying by $-1$ for each index flip. If any index is repeated it's zero. In some sense, the wedge product encodes that algebra so you can avoid using the antisymmetric symbol. It is also called the Levi Civita symbol in some contexts. By the way, check out my answer. $\endgroup$ – James S. Cook Aug 7 '13 at 3:19

I think I understand your question now. The wedge product does not produce the absolute value of the determinant. Instead, it produces the determinant itself in the following sense: if $A$ is a square matrix and $e_i$ is the usual standard basis for $\mathbb{R}^n$ then $$ Ae_1 \wedge Ae_2 \wedge \cdots \wedge Ae_n = det(A) e_1 \wedge e_2 \wedge \cdots \wedge e_n $$ This implicitly defines the determinant of $A$. One could reasonably define $det(A)$ by: $$ (Ae_1 \wedge Ae_2 \wedge \cdots \wedge Ae_n)(e^1,e^2, \dots ,e^n) = det(A) $$ where I replace $e_1$ with its cannonical double dual.

In any event, the formula with absolute value of the determinant is from the change of variables for multivariate integrals. Rather than try to explain in general, let's look at an example. Consider the integral of $f(x,y) = x^2+y^2$ on $1 \leq x^2+y^2 \leq 4$. This is an annulus, let's denote it $R \subseteq \mathbb{R}_{xy}^2$. We can calculate: $$ \iint_R (x^2+y^2) dx \, dy $$ Suppose we change variables to $\theta, r$. Note: $$ dx \wedge dy = (\cos \theta dr - r\sin\theta d\theta) \wedge (\sin \theta dr +r\cos\theta d\theta) = rdr \wedge d\theta = - rd\theta \wedge dr$$ Therefore, $dx \wedge dy$ is naturally replaced with $- rd\theta \wedge dr$. But, this is not what we use to change the integral to $\theta, r$ variables. Instead, $$ \iint_R (x^2+y^2) dx \, dy = \iint_S r^2 \frac{\partial (x,y)}{\partial(\theta, r)} d\theta dr $$ The Jacobian determinant is define to by: (in most calculus texts) $$ \frac{\partial (x,y)}{\partial(\theta, r)} = \bigg| det\left[ \begin{array}{cc} \partial x /\partial \theta & \partial x /\partial r \\ \partial y /\partial \theta & \partial y /\partial r \end{array}\right]\bigg| = | -r | = r$$ Thus, as $S = [0,2\pi] \times [1,2] \subseteq \mathbb{R}_{\theta r}^2$ $$ \iint_R (x^2+y^2) dx \, dy = \int_1^2\int_0^{2\pi} r^3 d\theta dr = \frac{15\pi}{2}.$$ Usually, I order the variables $r,\theta$ so the absolute value is not needed, but generally it is there to kill a sign for the area element (or more generally volume, hypervolume, whatver you want to call $dx_1 \dots dx^n$). This absolute value is the same which appears in the triple-scalar product formula: $$ vol = | \vec{A} \cdot (\vec{B} \times \vec{C})|$$ the absolute value is there to kill a minus sign, if the vectors are correctly oriented then the triple is right-handed which means the triple-scalar product is positive. However, if the triple $\vec{A},\vec{B},\vec{C}$ is left handed then $\vec{A} \cdot (\vec{B} \times \vec{C})<0$. Moreover, it should be noted that: $$ \vec{A} \cdot (\vec{B} \times \vec{C}) = det(\vec{A}|\vec{B}|\vec{C})$$ and the latter formula generalizes to arbitrary dimension.

In short, wedge products encode orientation, but ordinary area, volume etc... integrals are defined to give values independent of the ordering of the variables with which they are calculated. Area is positive. Volume is positive. This forces the absolute value.

  • $\begingroup$ Aha, got it~ Thank you James~ $\endgroup$ – 1LiterTears Aug 7 '13 at 15:13

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