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$\newcommand{\IP}[2]{\left\langle #1,#2 \right\rangle}$ $\newcommand{\de}{\delta}$

Let $M$ be a Riemannian manifold, and let $e_i$ be a local frame for $TM$. We also choose some fixed differential forms $\omega \in \Omega^k(M),\eta \in \Omega^{k+1}(M)$. Consider the following vector field:

$$ X=\IP{e^i \wedge \omega }{\eta}e_i \tag{1} $$

(The inner product on forms is the one induced by the metric on $M$).

I claim $X$ is well-defined, i.e does not depend on the chosen frame $e_i$.

Is there a "conceptual proof" for that?

I have a proof which is nothing but a routine calculation (see below), but I wonder if there is a nicer explanation for why this "magic" happens.

(BTW, this vector field, or more precisely its divergence, arises naturally when trying to compute a formula for the coderivative).

For completeness, here is my proof:

Let $f_i$ be a different frame for $TM$, and let $f^j,e^i$ be the dual bases of $f_j,e_i$ respectively. Write $$ f_i=A_i^{j}e_j,f^i=B_j^i e^j. $$ We claim $A=B^{-1}$. Indeed, $$ \de^i_k= f^i(f_k)=B_j^ie^j(A_k^se_s)=B_j^iA_k^s\de^j_s=\sum_j B_j^iA_k^j=(BA)_{ik}. $$ So, $$ \begin{split} \IP{f^i \wedge \omega }{\eta}f_i &=\IP{ B_j^i e^j \wedge \omega }{\eta}A_i^{k}e_k=A_i^{k}B_j^i\IP{ e^j \wedge \omega }{\eta}e_k=(AB)^k_j\IP{ e^j \wedge \omega }{\eta}e_k \\ &=\de^k_j\IP{ e^j \wedge \omega }{\eta}e_k= \IP{e^i \wedge \omega }{\eta}e_i. \end{split} $$

Edit:

Here is summary of Andreas and levap's suggestions:

Let $V$ be a real vector space.

Fix $\omega \in \Lambda^{k} V^*,\eta \in \Lambda^{k+1} V^*$.

Look at the bilinear map: $V^* \times V \to \Lambda^{k+1} V^* \otimes \Lambda^{k+1} V^* \otimes V$ defined by

$$(\hat{\omega}, v) \to \eta \otimes (\hat{\omega} \wedge \omega) \otimes v.$$

This map induces a linear map $\Phi:V^* \otimes V \to \Lambda^{k+1} V^* \otimes \Lambda^{k+1} V^* \otimes V$,

$$\Phi(\hat{\omega}\otimes v) := \eta \otimes (\hat{\omega} \wedge \omega) \otimes v.$$

Now, put an inner product on $\Lambda^{k+1} V^*$ (which can be induced by a product on $V$ in a natural way, or not, it does not matter). This product, which by definition is a bilinear map $\Lambda^{k+1} V^* \times \Lambda^{k+1} V^* \to \mathbb{R}$ induces a linear map $ \Lambda^{k+1} V^* \otimes \Lambda^{k+1} V^* \to \mathbb{R}$. By tensoring it with the identity map $\text{Id}_V$ we get

$$ \Psi: \Lambda^{k+1} V^* \otimes \Lambda^{k+1} V^* \otimes V \to \mathbb{R} \otimes V \cong V , \tag{2}$$ given by

$$ \Psi (\eta_1 \otimes \eta_2 \otimes v)=\IP{\eta_1}{\eta_2}v. \tag{3}$$

Now, note that for any basis $e_i$ for $V$, $$e^i \otimes e_i = \text{Id}_V, \tag{4}$$ where $e^i$ is the dual basis for $e_i$, and we have used the canonical identification $V^* \otimes V \cong \text{Hom}(V,V)$.

It is this fact that, which holds for any frame $e_i$, which lies in the heart of the "basis-independence" of the vector field defined in $(1)$.

Indeed,

$$ \Phi(\text{Id}_V)=\Phi(e^i \otimes e_i) = \eta \otimes (e^i \wedge \omega) \otimes e_i,$$ and

$$ \Psi \circ \Phi:\text{Hom}(V,V) \to V.$$

Finally, definition $(1)$ is nothing but setting

$$ X:=\Psi \circ \Phi (\text{Id}_V)=\Psi\big(\eta \otimes (e^i \wedge \omega) \otimes e_i\big)=\IP{\eta }{e^i \wedge \omega }e_i.$$

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  • $\begingroup$ You can consider the map $(\hat{\omega}, \hat{X}) \mapsto \left< \hat{\omega} \wedge \omega, \eta \right> \hat{X}$ where $\hat{\omega} \in \Omega^1(M), \hat{X} \in \mathcal{X}(M)$. This map is $C^{\infty}(M)$-bilinear and hence corresponds to a bundle map $\Phi \colon T^{*} M \otimes TM \rightarrow TM$. The vector field $X$ is just $\Phi(\operatorname{Id}_{TM})$ (under the identification of $T^{*}M \otimes TM \cong \operatorname{Hom}(TM, TM)$). From this description you can also see that the vector field $X$ is defined in local coordinates by your formula using arbitrary frames, not only $\endgroup$ – levap Dec 27 '17 at 14:51
  • $\begingroup$ orthonormal ones. $\endgroup$ – levap Dec 27 '17 at 14:52
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The conceptual explanation is that there is a bilinear map $\Lambda^k\mathbb R^{n*}\times\Lambda^{k+1}\mathbb R^{n*}\to\mathbb R^n$ which is $O(n)$-equivariant and thus defines a corresponding natural operation on Riemannian manifolds. Your computation actually shows equivariancy of this map (and the fact that things depend on a point plays no role in that computation). You can also see this without computation by observing that $\Lambda^k\mathbb R^{n*}\cong\Lambda^k\mathbb R^n$, there is a pairing $\Lambda^k\mathbb R^n\times \Lambda^{k+1}\mathbb R^{n*}\to\mathbb R^{n*}$ (by insertion, hence even $GL(n,\mathbb R)$-equivariant) and that $\mathbb R^{n*}\cong\mathbb R^n$. In fact it is only the last isomorphism that you express in terms of an orthonormal basis.

Edit: The pairing $\Lambda^k\mathbb R^{n*}\times\Lambda^k\mathbb R^n\to\mathbb R$ is indeed $GL(n,\mathbb R)$-equivariant with the group acting on both factors in the left hand side and trivially on $\mathbb R$. The computation in your answer does not work out since you have to the action on one factor wrong. What you need is $A\cdot (v_1\wedge\dots\wedge v_k)=(Av_1)\wedge\dots\wedge (Av_k)$ but interpreting $\eta$ as a multilinear map, the correct action is $(A\cdot\eta)(v_1,\dots,v_k)=\eta(A^{-1}v_1,\dots,A^{-1}v_k)$. From this the equivariancy is obvious.

Regarding the general construction of the map, my description was not strictly parallel to the formula for the map you have used (I overlooked that you were using a general frame and its dual rather than an orthonormal frame). Anyway, the description you use is better phrased as follows: There is a $GL(n,\mathbb R)$-equivariant map $\Lambda^k\mathbb R^{n*}\to \Lambda^{k+1}\mathbb R^{n*}\otimes\mathbb R^n$. This is given by first tensorizing with the identity and then applying the wedge-product. So in terms of dual bases $\{e_i\}$ for $\mathbb R^n$ and $\{e^i\}$ for $\mathbb R^{n*}$, this is given by $$ \omega\mapsto \textstyle\sum_ie^i\otimes \omega\otimes e_i\mapsto \textstyle\sum_ie^i\wedge\omega\otimes e_i. $$ (If you wonder about the succession of factors, this only means an overall sign, which does not influence equivariancy.) Tanking the tensor product with the identity, you get a map $\Lambda^{k+1}\mathbb R^{n*}\otimes \Lambda^k\mathbb R^{n*}\to \Lambda^{k+1}\mathbb R^{n*}\otimes \Lambda^{k+1}\mathbb R^{n*}\otimes\mathbb R^n$. Via the inner product, you get an $O(n)$-equivariant map from the first two factors in this tensor product to $\mathbb R$, and via the tensor product with the identity, this induces an $O(n)$-equivariant map $\Lambda^{k+1}\mathbb R^{n*}\otimes \Lambda^{k+1}\mathbb R^{n*}\otimes\mathbb R^n\to\mathbb R\otimes \mathbb R^n\cong \mathbb R^n$. The composition of this with the above gives the desired $O(n)$-equivariant map we were looking for.

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  • $\begingroup$ Thanks, this is interesting. I do have some questions on your interpretation though: (1) If I am not mistaken, the equivariancy of the first action $\Lambda^k\mathbb R^{n*}\times\Lambda^{k+1}\mathbb R^{n*}\to\mathbb R^n$ follows from the fact that the definition of the vector (field) does not depend on which orthonormal basis we choose, but in fact we have here something much stronger, namely total independent of the basis, orthonormal or not. (You can see details in my "answer" below)... $\endgroup$ – Asaf Shachar Dec 27 '17 at 8:33
  • $\begingroup$ (2) I am not sure exactly what do you mean by the $GL(n)$-equivariance? Perhaps you meant $GL(n)$ to act only on one of the factors? (if it acts on both there is no equivariance...see part 2 in my answer). I am also not sure about your last comment on "only in the last isomorphism we use orthonormal basis*; as I said, everything here works with arbitrary bases. $\endgroup$ – Asaf Shachar Dec 27 '17 at 8:33
  • $\begingroup$ @AsafShachar: I have edited my answer to address your comments. $\endgroup$ – Andreas Cap Dec 27 '17 at 10:02
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In the light of tensor calculus, $\IP{e^i \wedge \omega}{\eta}$ being a linear map $TM^* \to \bigsqcup_{p\in M} \mathbb R$ defines a tensor $\IP{\omega}{\eta}^i$. The upper index $i$ then cancels the lower index $i$ of $e_i$ resulting in a quantity that is invariant under change of basis.

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  • $\begingroup$ I think $\IP{e^i \wedge \omega}{\eta}$ is a real number, not a linear map. $\endgroup$ – Asaf Shachar Dec 27 '17 at 7:12
  • $\begingroup$ @AsafShachar. I meant that $u \mapsto \IP{u \wedge \omega}{\eta}$ is a linear map. $\endgroup$ – md2perpe Dec 27 '17 at 8:46
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I am writing some of the details of Andreas Cap's answer.

Part 1:

Define $F:\Lambda^k\mathbb R^{n*}\times\Lambda^{k+1}\mathbb R^{n*}\to\mathbb R^n$ by

$$ F(\omega,\eta)=\IP{e^i \wedge \omega }{\eta}e_i, \tag{1}$$

where $e_i$ is the standard basis of $\mathbb{R}^n$. Define an action of $O(n)$ on $\Lambda^k\mathbb R^{n*}$ by $(Q,\omega) \to Q\omega$, $Q \in O(n),\omega \in \Lambda^k\mathbb R^{n*}$, given by

$$ (Q \omega)(v_1,\dots,v_k)= \omega (Qv_1,\dots,Qv_k).$$ This action extends naturally to the product $\Lambda^k\mathbb R^{n*}\times\Lambda^{k+1} \mathbb R^{n*}$ by acting on both components.

Of course, $O(n)$ also acts on $\mathbb R^n$ in the natural way.

Lemma: $F$ is $O(n)$-equivariant.

Proof:

The proof follows from the fact that the definition of $F$ in $(1)$ gives the same map when replacing $e_i$ with any other orthonormal basis.

Indeed, suppose this basis-independence; then

$$F(Q\omega,Q\eta)=\IP{e^i \wedge Q\omega }{Q\eta}e_i=\IP{Qe^i \wedge Q\omega }{Q\eta}Qe_i=\IP{Q(e^i \wedge \omega) }{Q\eta}Qe_i=\IP{e^i \wedge \omega }{\eta}Qe_i=Q(\IP{e^i \wedge \omega }{\eta}e_i)=QF(\omega,\eta),$$

where we used the fact the $O(n)$-action on $\Lambda^k\mathbb R^{n*}$ is an action by isometries (i.e. preserve the norm on $\Lambda^k\mathbb R^{n*}$).

Part 2:

I am trying to understand Andreas's next remark. He claims there exist a $GL(n)$-equivariant map $$H:\Lambda^k\mathbb R^n\times \Lambda^{k+1}\mathbb R^{n*}\to\mathbb R^{n*},$$

defined by by insertion. This is the map $$\big( H(v_1 \wedge v_2 \wedge \dots \wedge v_k ,\eta)\big)(X) \to \eta(v_1 \wedge v_2 \wedge \dots \wedge v_k \wedge X).$$

Now, when trying to check $GL(n)$-equivariancy, I hit a problem: Let $A \in GL(n)$; then

$$ H(A(v_1 \wedge v_2 \wedge \dots \wedge v_k) ,A\eta)(X)=A\eta(Av_1 \wedge Av_2 \wedge \dots \wedge Av_k \wedge X)=$$

$$ \eta(A^2v_1 \wedge A^2v_2 \wedge \dots \wedge A^2v_k \wedge AX) \neq \eta(v_1 \wedge v_2 \wedge \dots \wedge v_k \wedge AX)=H(v_1 \wedge v_2 \wedge \dots \wedge v_k ,\eta)(AX)=\big( AH(v_1 \wedge v_2 \wedge \dots \wedge v_k ,\eta)\big)(X).$$

Perhaps $GL(n)$ only acts on one of the factors?

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