This is an old exam question which most of it I understand now due to the comments below. I still have two concerns.

Define a $p$-form on $GL(n)$ as follows. $$\Theta_p=tr(X^{-1}dX \wedge X^{-1}dX \wedge \cdots \wedge X^{-1}dX)$$ (i) Restrict $\Theta_3$ to $O(3)$. Writing the matrix $X$ with orthonormal column vectors $\mathbf{x_1, x_2,x_3}$, show that $X^{-1}dX$ is a skew symmetric matrix whose $i,j$th entry is $\mathbf{x}_i\cdot d\mathbf{x}_j$.

(ii) Show that at $X=I$, the forms $\mathbf{x_1} \cdot \mathbf{dx_2}, \mathbf{x_2 \cdot dx_3},$ and $\mathbf{x_3, \cdot dx_1}$ are linearly independent and $\Theta_3$ is nonzero.

(iii) Deduce that since $(AX)^{-1}d(AX)=X^{-1}dX$, $\Theta_3$ is non vanishing at all points.

(a) why can we just restrict $\Theta_3$ to $O(3)$? How do we know this form is still smooth? i.e. the induced map to $O(3) \rightarrow \wedge^*T^*O(3)$ is still smooth?

(b) Why does the equality in (iii) show $X^{-1}dX$ is left invariant?

I am also unsure what this means, since we are working with $1$-forms rather than vector fields. I suppose we are to show: $$(L_{A^{-1}}^*) X^{-1}dX_I = X^{-1}dX_A $$

  • $\begingroup$ The restriction is the pullback under the inclusion. It's always well defined. For (ii), be careful, the entries of the matrices don't commute anymore, so it's not clear that the cube of a skew-symmetric matrix is still skew-symmetric (and it's probably not, otherwise this question would be silly). For (iii), it's just saying that the form is left-invariant, and therefore if it is nonvanishing at one point, it's nonvanishing at all points. $\endgroup$ – jgon May 24 '19 at 20:34
  • $\begingroup$ @jgon, thanks i see what you mean from (ii), but I don't understand how showing the three forms are linearly independent shows $\Theta_3$ is nonzero. $\endgroup$ – W. Zhan May 24 '19 at 20:47

Yes, the wedge-product on $\Omega^\bullet(GL(n;\mathbb{R}))$ (or $\mathbb{C}$) and the matrix product on $\mathbb{R}^{n\times n}$ together defines a product on the tensor product of rings $\Omega^\bullet(GL(n;\mathbb{R}))\otimes_{\mathbb{R}}\mathbb{R}^{n\times n}$ which is denoted also by $\wedge$, and your formula is correct. Here $X\colon GL(n;\mathbb{R})\to\mathbb{R}^{n\times n}$ is just the inclusion.

(1) $\Theta_p$ is a $p$-form on $GL(n)$ so it restricts (actually pulls back) to $O(n)\subset GL(n)$ under the usual inclusion. Since $X^{-1}=X^T$ for $X\in O(3)$, the form $X^{-1}\,\mathrm{d}X$ is $$ X^{-1}\,\mathrm{d}X=(\mathbf{x}_1\,\mathbf{x}_2\,\mathbf{x}_3)^T(\mathrm{d}\mathbf{x}_1\,\mathrm{d}\mathbf{x}_2\,\mathrm{d}\mathbf{x}_3)=(\mathbf{x}_i^T\mathrm{d}\mathbf{x}_j)=(\mathbf{x}_i\cdot\mathrm{d}\mathbf{x}_j). $$ To show this is skew-symmetric, remember $(X^T)\,\mathrm{d}X+\mathrm{d}(X^T)X=0$ on $O(3)$. Taking the $(i,j)$th component yields: $$ 0=\sum_k(X^T)_{ik}(\mathrm{d}X)_{kj}+(\mathrm{d}X^T)_{ik}X_{kj} =\sum_kX_{ki}\,\mathrm{d}X_{kj}+X_{kj}\,\mathrm{d}X_{ki} =\mathbf{x}_i\cdot\mathrm{d}\mathbf{x}_j+\mathbf{x}_j\cdot\mathrm{d}\mathbf{x}_i. $$

(2) At identity, $\mathbf{x}_i\cdot\mathrm{d}\mathbf{x}_j=\mathrm{d}X_{ij}$, so the three forms are linearly independent. We have $\Theta_3$ is $$ \sum_{i,j,k}(\mathbf{x}_i\cdot\mathrm{d}\mathbf{x}_j)\wedge(\mathbf{x}_j\cdot\mathrm{d}\mathbf{x}_k)\wedge(\mathbf{x}_k\cdot\mathrm{d}\mathbf{x}_i) $$ the summands are nonzero only if $i,j,k$ pairwise distinct, so this is sum over permutations of $1,2,3$. Show that $(\mathbf{x}_i\cdot\mathrm{d}\mathbf{x}_j)\wedge(\mathbf{x}_j\cdot\mathrm{d}\mathbf{x}_k)\wedge(\mathbf{x}_k\cdot\mathrm{d}\mathbf{x}_i)$ is invariant under all permutations (which you should be able to do). So we conclude, at identity, $$ \Theta_3=6(\mathbf{x}_1\cdot\mathrm{d}\mathbf{x}_2)\wedge(\mathbf{x}_2\cdot\mathrm{d}\mathbf{x}_3)\wedge(\mathbf{x}_3\cdot\mathrm{d}\mathbf{x}_1) \neq 0. $$

(3) Since $(AX)^{-1}\,\mathrm{d}(AX)=X^{-1}A^{-1}A\,\mathrm{d}X=X^{-1}\,\mathrm{d}X$, the matrix of $1$-forms $X^{-1}\,\mathrm{d}X$ is left-invariant (on $GL(n)$). Hence $\Theta_3$ is invariant on $GL(3)$ and hence on $O(3)$. Together with (2), this shows $\Theta_3$ must be nonvanishing everywhere on $O(3)$.

  • $\begingroup$ Thank you so much, I have updated my question, and confusion on invariance. Hope you can have a look at it. $\endgroup$ – W. Zhan May 26 '19 at 6:17
  • $\begingroup$ Yes, $(AX)^{-1}\,\mathrm{d}(AX)$ is the pullback of $X^{-1}\,\mathrm{d}X$ under $L_{A^{-1}}$. The proof is a one-liner: note that $(AX)^{-1}\,\mathrm{d}(AX)$ is evaluating $X^{-1}\,\mathrm{d}X$ at $AX$, so is the pullback of a left translation by an element that goes from $AX$ back to $X$, i.e., $A^{-1}$. $\endgroup$ – user10354138 May 26 '19 at 7:47
  • $\begingroup$ As for (a), the obvious $O(n)\hookrightarrow GL(n)$ is a (smooth) immersion (in fact embedding), so you can pull the smooth form $\Theta_3$ on $GL(3)$ back to a smooth form on $O(3)$. $\endgroup$ – user10354138 May 26 '19 at 7:55

HINT for (ii): The tangent space to O(3) at the identity matrix consists of all skew-symmetric matrices. If $A$ is in that tangent space, show that $(\mathbf x_i\cdot d\mathbf x_j)(A) = a_{ij}$.

Then recall that forms $\theta_1,\dots,\theta_k$ are linearly independent if and only if $\theta_1\wedge\dots\wedge\theta_k\ne 0$.


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