Concrete Example Illustrating the Interior Product

Question

Let $V$ be a finite-dimensional vector space, let $v \in V$ and let $\omega$ be an alternating $k$-tensor on $V$, i.e., $\omega \in \Lambda^{k}(V)$. Then, the interior product of $v$ with $w$, denoted by $i_{v}$, is a mapping $$ i_{v}:\Lambda^{k}(V)\rightarrow \Lambda^{k-1}(V) $$ determined by

$$ (i_v \omega)(v_1, \dots, v_{k-1}) = \omega(v, v_1, \dots, v_{k-1}). $$

My understanding of this, which is probably far from complete, is that the interior product basically provides a mechanism to produce a $k-1$-tensor from a $k$ tensor relative to some fixed vector $v$. I'm trying to understand however what the interior product actually means and how it is used in practice. Therefore, my question is, Can anyone provide example(s) illustrating computations and/or physical examples that will shed light on its purpose?

Also, the interior product seems to be somewhat (inversely?) related to the exterior product in that an exterior product takes a $p$-tensor and a $q$ tensor and makes a $p+q$ tensor and therefore is an "expansion". The interior product, on the other hand, is a contraction but always produces a tensor of degree one less than you started out with. So, secondly, What is the precise relation between the interior and exterior products?

Unfortunately, the Wikipedia page is of little help here and I can't find a reference that clearly explains these things.

Answer 1

Let me give another illustration of how the interior and exterior products are related. This particular case, however, works not on differential geometry, but requires Riemannian geometry.

Given a metric $g$, denote by $\langle,\rangle$ the extension of its inner (not interior) product to forms. The metric $g$ induces an identification between the vector space $V$ and its dual $V^*$, via the operators $v\mapsto v^\flat$, where

$$ v^\flat(w) = \langle v,w\rangle $$

($v^\flat \in V^*$ is a linear functional on $V$, and here we define it by its action on $w\in V$)

Then we have the nice property for $\eta\in\Lambda^{k-1}(V),\tau \in \Lambda^k(V)$, and $v\in V$ that

$$ \langle v^\flat \wedge \eta, \tau\rangle = \langle \eta,(i_v)\tau \rangle $$

showing how the interior and exterior products are actually adjoint with respect to the metric inner product.

A similar statement can be made by appealing to the Hodge-star operator associated to a Riemannian metric. Up to a constant multiplier $C$ (whose form depends a bit on your conventions, and which depends on the dimension and the degree of the forms), you have that

$$ (i_x)\tau = C *(x^\flat\wedge *\tau) $$

where $*$ is the Hodge star operator.

Answer 2

Lie derivative, of course! It's even mentioned in the interior product wiki page. As for the relation to the exterior product, IMO you should look more on the exterior derivative for comparison.

But if you insist, for any $1$-form $\alpha$ and $k$-form $\omega$ and any vector $x$ we have $i_x (\alpha \wedge \omega) = \alpha(x) \omega$.

UPD: from purely algebraic standpoint, it is useful to consider Lie coalgebra structure on $V^*$: a linear mapping $\mathrm{d}: V^* \to \bigwedge^2 V^*$ extended to a graded anti-derivation. It is easy to see that by defining $\omega([x, y]) = \mathrm{d}\omega(x, y)$ we get Lie algebra structure on $V$.

So, interesting things must arise when we extend some $f: \bigwedge^2 V^* \to V^*$ to a graded antiderivation, right? Well, I can't answer that question, but I'm pretty sure that was the motivation :)

Answer 3

I don't think this will completely satisfy your questions, but I think the interior product is a neat way to induce orientations. To give an orientation on an $n$-manifold with boundary $M$ is the same as giving a nowhere-vanishing $n$-form $\Omega$. If $H \subset M$ is a hypersurface and $N$ is a transverse vector field along $H$ (so $N\colon H \to TM$, such that $N_x \in T_xM$ and $T_xM = N_x + T_xH$ for $x \in H$), then $i_N\Omega$ restricts to an orientation form on $H$. If $H = \partial M$, then taking $N$ to be an outward-pointing vector field along $\partial M$ gives the usual orientation used in Stokes's theorem.

I don't have my copy with me, but a lot of this should be in Lee's Introduction to Smooth Manifolds.

Answer 4

Here is a partial answer to my own question, specifically the part asking for a concrete example illustrating computational aspects of the interior product.

Let the vector space in question be $\mathbb{R}^3$ endowed with the ordered basis $(e_1, e_2, e_3)$ and let $e^1, e^2, e^3$ be the relative cobasis. Here, we can think of the cobasis as just ordinary vectors that satisfy $e^i(e_j) = \delta^i_j$ or consider them as linear functionals on the dual space $(\mathbb{R}^n)^*$ that satisfy the same relations.

Now, suppose $\omega \in \Lambda^{2}(\mathbb{R}^3)$ is given by $\omega = e^1 \wedge e^2$ and let $v = e_1$. Then, for any vector $x \in \mathbb{R}^3$ we can then compute the interior product as follows:

$$ (i_v \omega)(x) = \omega(v, x) = (e^1 \wedge e^2)(e_1, x) = e^1(e_1)e^2(x) - e^1(x)e^2(e_1) = e^2(x) $$

Therefore $i_v \omega = e^2$

Next, keep the same $\omega$ but let $v = e_2$. Then,

$$ (i_v \omega)(x) = (e^1 \wedge e^2)(e_2, x) = e^1(e_2)e^2(x) - e^1(x)e^2(e_2) = -e^1(x) $$

So $i_v \omega = -e^1$

Finally, it is also easy to see by inspection that if $v =e_3$ then $i_v \omega = 0$

Computations for other values of $\omega$ proceed similarly.

Answer 5

$\def\int{\operatorname{int}} \def\ext{\operatorname{ext}} \def\tsr{\operatorname{tsr}}$This post will answer the OP's question “what is the precise relation between the interior and exterior products?” It is Proposition 17. This result is the “nice property” that Willie Wong's explained in his answer, but without choosing a metric $g$. We will actually prove Willie Wong's formula in section 3 of this post by choosing a metric and bringing the situation to the general case. The good thing about all of this is that it is only linear algebra, and there is no smooth manifold theory. I've been unable to find the statement of Proposition 17 in any book, so I would be grateful if anyone could give a reference in case they know some.

1. Pairings and dualities over a field

Proposition 17 shows that “$\ext_v$ is the adjoint of $\int_v$ with respect to the pairing $\langle -, -\rangle :\Lambda V\times \Lambda V^* \to k$.” Before arriving to this result, we are going to dive into the theory of pairings and dualities over a field, which can be regarded as a subtheory of linear algebra. We will build enough theory to prove that such an adjoint for $\ext_v$ is unique. We won't directly use all the theory we are going to develop, but I think that it is instructing to see this minimal amount to put things in context.

Definition 1. As defined in Wikipedia, a pairing over a field $k$, or a $k$-pairing, is a triple $(V,W,b)$, where $V$ and $W$ are $k$-vector spaces and $b:V\times W\to k$ is a bilinear map. From a pairing we get induced linear maps \begin{align*} L_b:V&\to W^* \qquad\qquad R_b:W\to V^*\\ v&\mapsto b(v,-), \qquad\quad\quad \,\; w\mapsto b(-,w). \end{align*} The pairing $(V,W,b)$ is said to be non-degenerate if $L_b$ and $R_b$ are both injective. A dual system, dual pair or duality over a field $k$, or a $k$-duality, is a non-degenerate pairing over $k$.

The duality $(V,W,b)$ is finite-dimensional if $V$ and $W$ are finite-dimensional.

A morphism of $k$-pairings $\varphi:(V_1,V_2,b)\to (V_1',V_2',b')$ is a pair $\varphi=(\varphi_1,\varphi_2)$, where $\varphi_i:V_i\to V_i'$ is a $k$-linear map, for $i=1,2$, such that $b'(\varphi_1(v_1),\varphi_2(v_2))=b(v_1,v_2)$ for all $v_1\in V_1$, $v_2\in V_2$. A morphism of $k$-dualities is a just a morphism of $k$-pairings.

Remarks 2.

1. The $k$-pairings form a category $\mathsf{Prng}_k$ which has the category of $k$-dualities $\mathsf{Dlty}_k$ as a full subcategory.

2. The isomorphism of $k$-pairings $(V,W,b)\cong(W,V,b')$, where $b'(w,v)=b(v,w)$, is natural.

3. If $(V,W,b)$ is a finite-dimensional $k$-duality, then $\dim V=\dim W$, for injectivity of $L_b$ and of $R_b$ implies $\dim V\leq\dim W^*=\dim W$ and $\dim W\leq\dim V^*=\dim V$.

Examples 3.

1. If $k=\mathbb{R}$, then from a real inner product space $(V,\langle -,-\rangle)$ we obtain a real duality $(V,V,\langle -,-\rangle)$.

2. The $k$-pairing $\langle -,-\rangle:V\times V^*\to k$ defined by $\langle v,f\rangle = f(v)$ is a non-degenerate: On the one hand, $\langle -,f\rangle = f$; whereas on the other hand if $0=\langle v,-\rangle$, then $f(v)=0$ for all $f\in V^*$. Thus, if we take a basis $\{e_i\}$ of $V$, and we consider the associated linearly independent system $\{e_i^*\}\subset V^*$ (which is a basis if $\dim V<+\infty$), writing $v=v^ie_i$, we get $0=e_i^*(v)=v^i$, so $v=0$. For this reason, $(V,V^*,\langle -, -\rangle)$ or $(V^*,V,\langle -, -\rangle)$ is called the canonical $k$-pairing of the vector space $V$.

3. Let $V$ a $k$-vector space. Then the pair of maps $\operatorname{ev}:V\to V^{**}$, where $\operatorname{ev}(v)(f)=f(v)$, and $\operatorname{id}:V^*\to V^*$ form a natural morphism of canonical $k$-pairings $(V,V^*,\langle -,-\rangle )\to (V^{**},V^*,\langle -, -\rangle)$. For $V$ finite-dimensional, this is a natural isomorphism.

Definition 4. A $k$-subpairing of a $k$-pairing $(V,W,b)$ is a morphism $(\varphi,\psi):(V',W',b')\to(V,W,b)$ of pairings such that $\varphi$ and $\psi$ are injective. If $(V',W',b')$ is non-degenerate, it is called a $k$-subduality of $(V,W,b)$.

The proof of the following lemma is left as an exercise.

Lemma 5. Let $(V,W,b)$ and $(V',W',b')$ be $k$-pairings and let $\varphi:V\to V'$ and $\psi:W\to W'$ be linear maps. Then $(\varphi,\psi)$ is a morphism of $k$-pairings $(V,W,b)\to(V',W',b')$ if and only if the following diagrams commute: $$\require{AMScd} \begin{CD} V @>{L_b}>> W^* @. \qquad @. W @>{R_b}>> V^*\\ @V{\varphi}VV @AA{\psi^*}A @. @V{\psi}VV @AA{\varphi^*}A \\ V' @>>{L_{b'}}> (W')^* @. @. W' @>>{R_{b'}}> (V')^* \end{CD}$$ and if and only if at least one of them commutes.

Corollary 6. Let $(V,W,b)$, $(V',W',b')$ be isomorphic dualities. Then $L_b$ (resp., $R_b$) is injective if and only if $L_{b'}$ (resp., $R_{b'}$) is. In particular, $\mathsf{Dlty}_k$ is a replete subcategory of $\mathsf{Prng}_k$, meaning that two isomorphic $k$-pairings are $k$-dualities if and only if one of them is.

Proof. Immediate by inspection of the last commutative diagrams, since a linear map $T:A\to B$ being an iso implies, by functoriality, that the dual map $T^*:B^*\to A^*$ is an iso. $\square$

There is a simple situation in which for verifying that a pairing $(V,W,b)$ is non-degenerate is suffices to verify that at least one of $L_b$ or $R_b$ is injective.

Lemma 7. Let $(V,W,b)$ be a finite-dimensional $k$-pairing with $\dim V=\dim W$. Then $L_b$ and $R_b$ are injective if and only if one of them is.

Proof. By last Corollary, we may assume that $V=W$, since if we pick an isomorphism $\varphi:W\cong V$, then $(1_V,\varphi):(V,W,b)\to (V,V,b')$ is an isomorphism of pairings, where $b'(v_1,v_2)=b(v_1,\varphi^{-1}(v_2))$, for $v_i\in V$.

The following diagrams commute: $$\require{AMScd} \begin{CD} V @>{L_b}>> V^* @. \qquad @. V @>{R_b}>> V^*\\ @V{\operatorname{ev}}VV @| @. @V{\operatorname{ev}}VV @| \\ V^{**} @>>{R_{b}^*}> V^* @. @. V^{**} @>>{L_{b}^*}> V^* \end{CD}$$ where $\operatorname{ev}(v)(f)=f(v)$, for $v\in V$, $f\in V^*$. The map $\operatorname{ev}$ is an isomorphism (always true in the finite-dimensional case), and $L_b$ (resp., $R_b$) being injective implies it is bijective, by dimensionality. Thus, the claim follows since the dual functor reflects isomorphisms in finite-dimensional spaces ($T^*$ being an iso for a linear map $T:A\to B$ between finite-dimensional spaces implies $T$ is an iso). $\square$

Definition 8. Let $(V,W, b)$, $(V',W',b')$ be $k$-pairings, and $A:V\to V'$, $B:W'\to W$ be linear operators. With respect to the pairings $b$ and $b'$, we say that $A$ is a left adjoint or a left transpose to $B$, or that $B$ is a right adjoint or a right transpose to $A$ if $$ b'(Av,w')=b(v,Bw') $$ for all $v\in V$, $w'\in W'$. If there is no place for confusion, sometimes we will just say $A$ is an adjoint or transpose for $B$ or $B$ is an adjoint or transpose for $A$. We will also say that the tuple $(A,B,b,b')$ is an adjunction. In the case where $(V,W, b)=(V',W',b')$, we will also say that the triple $(A,B,b)$ is an adjunction.

Lemma 9. Let $(V,W,b)$, $(V',W',b')$ be $k$-pairings, and suppose that $R_b$ is injective (resp., suppose that $L_{b'}$ is injective). If a linear map $V\to V'$ (resp., a linear map $W'\to W$) has an right adjoint (resp., a left adjoint), then it is unique.

Proof. Suppose $B,\widetilde{B}:W'\to W$ are right adjoints for $A:V\to V'$, and let $w'\in W'$. Then, for all $v\in V$, $$ b(v,Bw')=b'(Av,w')=b(v,\widetilde{B}w'), $$ so $0=b(v,Bw-\widetilde{B}w)$. Since $R_b$ is injective, this implies $Bw=\widetilde{B}w$. Hence $B=\widetilde{B}$. Uniqueness of left adjoints is proven analogously. $\square$

Examples 10.

1. If $(V,V^*,\langle -,-\rangle)$ and $(W,W^*,\langle -,-\rangle)$ are the canonical $k$-pairings of the vector spaces $V$ and $W$, then a linear map $F:V\to W$ always has a right transpose, namely, $F^*$, since $$ \langle v,F^*\omega\rangle=\langle Fv,\omega\rangle, $$ for $v\in V$ and $\omega\in W^*$.

2. Let $V=k^n$ and suppose $\langle -, -\rangle : k^n\times k^n\to k$ is the standard bilinear form, $\langle v^ie_i,w^je_j\rangle = \sum_{i=1}^n v^iw^i$. For any matrix $A\in k^{n\times n}$, the transpose operator of $A$ (right or left) is given by the transpose matrix.

Lemma 11. Let $(V,W,b)$, $(V',W',b')$ be $k$-dualities and let $A:V\to V'$ be a linear map. A sufficient condition for the existence of a unique right adjoint of $A$ is that $R_b$ is injective and that $b'(A(-),W')\subset b(-,W)$, where \begin{align*} b'(A(-),W')&=\{b'(A(-),w')\mid w'\in W'\},\\ b(-,W)&=\{b(-,w)\mid w\in W\}, \end{align*} are subsets of $V^*$. Analogously, if $B:W'\to W$ is a linear map, a sufficient condition for the existence of a unique left adjoint of $B$ is that $L_{b'}$ is injective and $b(V,B(-))\subset b'(V',-)$.

Proof. Uniqueness follows from last lemma. We show existence of the right adjoint of $A$. Pick a $w'\in W'$. Then there is a (unique since $R_b$ is injective) $w\in W$ such that $b'(A(-),w')=b(-,w)$. Define $A^t(w')=w$. Then clearly $$ b'(Av,w')=b(v,A^tw') $$ for all $v\in V$. It remains to verify that $A^t:W'\to W$ is linear. Let $w_1',w_2'\in W'$. Then, for $v\in V$, we have \begin{align*} b(v,A^t(w_1'+w_2')) &=b'(Av,w_1'+w_2')\\ &=b'(Av,w_1')+b'(Av,w_2')\\ &=b(v,A^tw_1')+b(v,A^tw_2')\\ &=b(v,A^tw_1'+A^tw_2'). \end{align*} Thus, since $R_b$ is injective, it follows that $A^t(w_1'+w_2')=A^tw_1'+A^tw_2'$, so $A^t$ is additive. Homogeneity of $A^t$, i.e., $A^t(\lambda w')=\lambda A^tw'$, is proven similarly. Existence of the left adjoint of $B$ is done similarly. $\square$

Corollary 12. For finite-dimensional $k$-dualities, unique adjoints always exist. Explicitly, if $(V,W,b)$ and $(V',W',b')$ are finite-dimensional $k$-dualities, then, with respect to $(b,b')$, any linear map $V\to V'$ has a unique right adjoint and any linear map $W'\to W$ has a unique left adjoint.

Proof. This is because $R_b$ and $L_{b'}$ being injective implies they are isos in the finite-dimensional case, so the conditions $b'(A(-),W')\subset b(-,W)=V^*$ and $b(V,B(-))\subset b'(V',-)=(W')^*$ from last lemma are always satisfied. $\square$

Lemma 13. Let \begin{align*} (\varphi,\psi) &:(V_1,W_1,b_1) \to (V_2,W_2,b_2),\\ (\varphi',\psi')&:(V_1',W_1',b_1')\to (V_2',W_2',b_2'), \end{align*} be isomorphisms of $k$-pairings. If $(A_1,B_1,b_1,b_1')$ is an adjunction, then so is $(A_2,B_2,b_2,b_2')$, where \begin{align*} A_2&=\varphi' A_1\varphi^{-1},\\ B_2&=\psi B_1(\psi')^{-1}. \end{align*}

The proof is left as an exercise.

Lemma 14. Let $(V_i,W_i,b_i)$, $i\in I$, be a family $k$-pairings. The $k$-pairing $\bigoplus_i(V_i,W_i,b_i):=(\bigoplus_i V_i,\bigoplus_i W_i,\bigoplus_i b_i)$, defined as $\left(\bigoplus_i b_i\right)(\sum_iv_i,\sum_iw_i)=\sum_ib_i(v_i,w_i)$, is non-degenerate if and only if $(V_i,W_i,b_i)$ is non-degenerate for all $i\in I$.

Beware that this construction is not a coproduct in the category of pairings. It's neither a product for $I$ finite.

Proof. ($\Rightarrow$) Suppose that $v\in V_j$ is such that $b_j(v,w)=0$ for all $w\in W_j$. Then $(\bigoplus_i b_i)(v,w)=b_j(v,w_j)=0$ for all $w\in\bigoplus_i W_i$, where $w_j$ is the $j$-th component of $w$. Thus $v=0$, and $L_{b_j}$ is injective. Proving that $R_{b_j}$ is injective is analogous.

($\Leftarrow$) Now suppose $v\in \bigoplus_i V_i$ is such that $\left(\bigoplus_i b_i\right)(v,w)=0$ for all $w\in\bigoplus_i W_i$. In particular, this implies $0=\left(\bigoplus_i b_i\right)(v,w)=b_j(v_j,w)$ for all $w\in W_j$, where $v_j$ is the $j$-th component of $v$. Thus $v_j=0$ and therefore $v=0$, so $L_{\bigoplus_i b_i}$ is injective. Proving that $R_{\bigoplus_i b_i}$ is injective is analogous. $\square$

Lemma 15. Suppose $\operatorname{char} k\neq 2$ and let $(V_i,W_i,b_i)$ be finite-dimensional $k$-pairings, for $i=1,\dots,n$. The universal property of the tensor product induces a $k$-pairing $\bigotimes_i(V_i,W_i,b_i)=\left(\bigotimes_iV_i,\bigotimes_iW_i,\bigotimes_ib_i\right)$ by $(\bigotimes_ib_i)(v_1\otimes\cdots\otimes v_n,w_1\otimes\cdots\otimes w_n)=b_1(v_1,w_1)\cdots b_n(v_n,w_n)$ that is non-degenerate if $(V_i,W_i,b_i)$ is non-degenerate for all $i$. The converse is true if $V_i,W_i\neq \{0\}$.

Proof. We prove it for $n=2$. The proof for the general case is the same.

We prove the converse first. Denote $(V_1,W_1,b_1)=(V,W,b)$ and $(V_2,W_2,b_2)=(V',W',b')$ and suppose that $b\otimes b'$ is non-degenerate. Suppose $v\in V$ is such that $b(v,w)=0$ for all $w\in W$. Since $V'\neq\{0\}$, pick $v'\in V'\setminus\{0\}$. Then $$ (b\otimes b')(v\otimes v',w\otimes w')=b(v,w)b'(v',w')=0 $$ for all $w\in W$ and $w'\in W'$. Thus $v\otimes v'=0$, since $b\otimes b'$ is non-degenerate. Hence, since $v'\neq 0$, $v=0$ and $L_b$ is injective. Showing injectivity of $R_b,L_{b'},R_{b'}$ is analogous.

To prove now $(\Rightarrow)$, note that we can assume $V_i=W_i$. Suppose then $(V,V,b)$ and $(W,W,b')$ to be finite-dimensional $k$-dualities and let's show that $(V\otimes W,V\otimes W,b\otimes b')$ is non-degenerate. Pick a $b$-orthonormal basis $\{e_i\}$ of $V$ and a $b'$-orthonormal basis $\{f_j\}$ of $W$ (for this we need $\operatorname{char}k\neq 2$). Then $b\otimes b'$ is non-degenerate since $$ (b\otimes b')(e_i\otimes f_j,e_\ell\otimes f_m)=\delta_{i\ell}\delta_{jm}.\qquad\qquad\square $$

2. The adjoints of $\tsr_v$ and $\ext_v$

In the following, $V$ will always be a finite-dimensional vector space over a field $k$ of characteristic $0$. Denote $\Phi^p:T^pV^*\cong(T^pV)^*$ to the canonical isomorphism and write $$ \tag{1}\label{eq2} \langle t,\omega\rangle\overset{\mathrm{def}}{=}\langle t,\Phi^p(\omega)\rangle, $$ for $t\in T^p V$, $\omega\in T^pV^*$. By Corollary 6, this defines a $k$-duality $$ \tag{2}\label{p_dlty} \langle -,-\rangle: T^pV\times T^pV^*\to k, $$ since we have an isomorphism of $k$-pairings $(1_{T^pV},\Phi^p):(T^pV,T^pV^*,\langle -,-\rangle)\cong(T^pV,(T^pV)^*,\langle -,-\rangle)$. For the pairing \eqref{p_dlty}, one can verify that $$ \langle v_1\otimes\cdots\otimes v_n,v_1^*\otimes\cdots\otimes v_n^*\rangle =\langle v_1,v_1^*\rangle\cdots\langle v_n,v_n^*\rangle $$ for any $v_i\in V$ and $v_i^*\in V^*$.

By Lemma 14, the $k$-dualities \eqref{p_dlty} extend to a $k$-duality $$ \tag{3}\label{TV_dlty} \langle -,-\rangle: TV\times TV^*\to k. $$

For $t\in TV$, define an operator \begin{align*} \tsr_t:TV&\to TV\\ s&\mapsto t\otimes s. \end{align*} For $v\in V$ and $p\geq 1$, define \begin{align*} \tag{4}\label{eq:int_v} \int_v: T^pV^*\to T^{p-1}V^* \end{align*} as $(\int_v\omega)(v_2,\dots,v_p)=\omega(v,v_2,\dots,v_p)$, for $\omega\in T^pV^*$ (for $p=1$ we understand $\int_v\omega=\omega(v)$). When $p=0$, define $\int_v:T^0V\to \{0\}$. Using the canonical isomorphisms $T^qV^*\cong(T^qV)^*$, the resulting map $\int_v:(T^pV)^*\to(T^{p-1}V)^*$ with $p\geq 1$ is expressed as $\int_v\omega=\omega\circ\tsr_v:T^{p-1}V\to k$, for $\omega\in(T^pV)^*$. In other words, $$ \tag{5}\label{eq} \langle t,\int_v\omega\rangle=\langle\tsr_vt,\omega\rangle, $$ for $t\in T^{p-1}V$ and $\omega\in (T^pV)^*$. Thus, \eqref{eq} also holds for $t\in T^{p-1}V$ and $\omega\in T^p V^*$. As the linear maps \eqref{eq:int_v} assemble into an operator $$ \label{eq:int_v_all}\tag{6} \int_v:TV^*\to TV^*, $$ relation \eqref{eq} also holds for every $t\in TV$ and $\omega\in TV^*$. Equivalently, $\int_v$ is the right adjoint of $\tsr_v$ with respect to the $k$-duality \eqref{TV_dlty} (Lemma 9).

Let $\psi:\{0,1,2,\dots\}\to\{1,2,3,\dots\}$ be any function such that $\psi(0)=\psi(1)=1$. Recall that given $\alpha\in\Lambda^pV$ and $\beta\in\Lambda^qV$, one defines the exterior product $\alpha\wedge_\psi\beta\in\Lambda^{p+q}V$ as $$ \alpha\wedge_\psi\beta=\frac{\psi(p+q)}{\psi(p)\psi(q)}A_{p+q}(\alpha\otimes\beta), $$ where \begin{align*} A_p:T^pV&\longrightarrow\Lambda^pV\\ v_1\otimes\cdots\otimes v_p &\longmapsto\frac{1}{p!}\sum_{\sigma\in S_p}(\operatorname{sgn}\sigma) v_{\sigma(1)}\otimes\cdots \otimes v_{\sigma(p)} \end{align*} is the anti-symmetrization operator. Some authors define $\psi(n)=1$ to be constant, whereas others use the convention $\psi(n)=n!$. In general, one can check that $\wedge_\psi$ is associative and that $$ \alpha_1\wedge_\psi\cdots\wedge_\psi\alpha_n =\frac{\psi(p_1+\cdots+p_n)}{\psi(p_1)\cdots\psi(p_n)}A_{p_1+\dots+p_n}(\alpha_1\otimes\cdots\otimes\alpha_n), $$ for $\alpha_i\in\Lambda^{p_i}V$. Henceforth we shall denote $\wedge=\wedge_\psi$ and keep the function $\psi$ to be general so the following arguments are valid for all possible normalizations of the exterior product that show up in the literature.

Exercise 16. Given vectors $v_1,\dots,v_n\in V$ and covectors $v_1^*,\dots,v_n^*\in V^*$, show \begin{align*} \langle v_1\wedge\cdots\wedge v_n,v_1^*\otimes\cdots\otimes v_n^*\rangle &=\frac{\psi(n)}{n!}\det(\langle v_i,v_j^*\rangle)=\langle v_1\otimes\cdots\otimes v_n,v_1^*\wedge\cdots\wedge v_n^*\rangle,\\ \langle v_1\wedge\cdots\wedge v_n,v_1^*\wedge\cdots\wedge v_n^*\rangle &=\frac{\psi(n)^2}{n!}\det(\langle v_i,v_j^*\rangle). \end{align*} In particular, one has $$ \langle v_1\wedge\cdots\wedge v_n,\omega\rangle =\psi(n)\langle v_1\otimes\cdots\otimes v_n,\omega\rangle, $$ for $\omega\in\Lambda^nV^*$.

Since $\int_v(\Lambda^pV^*)\subset\Lambda^{p-1}V^*$ (check), the map \eqref{eq:int_v_all} restricts to a linear map $$ \int_v:\Lambda V^*\to\Lambda V^*. $$

For $\eta\in\Lambda V$, define \begin{align*} \ext_\eta:\Lambda V&\to\Lambda V\\ \nu&\mapsto\eta\wedge\nu. \end{align*}

If we pass the subspace $\Lambda^pV^*\subset T^pV^*$ through the canonical isomorphism $\Phi^p:T^pV^*\cong(T^pV)^*$, we get an isomorphism $\Lambda^pV^*\cong(\Lambda^pV)^*\subset(T^pV)^*$. Thus, the canonical duality of $\Lambda^pV$, via \eqref{eq2}, induces a duality $(\Lambda^pV,\Lambda^pV^*,\langle -, -\rangle)$ (Corollary 6). By Lemma 14, these assemble into a duality $$ \tag{7}\label{dlty_ext_pow} \langle -, -\rangle :\Lambda V\times \Lambda V^* \to k, $$ which is a subduality of \eqref{TV_dlty}. Indeed, the pairings $\langle -, -\rangle_{\Lambda V,\Lambda V^*}$, $\langle -, -\rangle_{T V,T V^*}$ are induced by the canonical pairings on $\Lambda^pV$ and $T^p V$, and in these last ones the former is a subpairing of the latter.

Proposition 17. For $\omega\in(\Lambda^pV)^*$ and $v\in V$, we have $$ \int_v\omega=\omega\circ\ext_v:\Lambda^{p-1}V\to k. $$ Equivalently, $$ \tag{8}\label{ext_adj_int} \langle\eta,\int_v\omega\rangle =\langle\ext_v\eta,\omega\rangle $$ for $\eta\in\Lambda^{p-1}V$. This last equality is also valid when considering $\omega\in\Lambda^pV^*$, and, hence, also valid for all $\eta\in\Lambda V$ and $\omega\in\Lambda V^*$.

In other words, $\int_v$ is the right adjoint of $\ext_v$ with respect to the duality \eqref{dlty_ext_pow} (Lemma 9).

Proof. It suffices to show \eqref{ext_adj_int} for a system of generators of $\Lambda^p V^*$ and of $\Lambda^{p-1}V$. Let $v_1,\dots,v_p\in V$ be vectors and $v_1^*,\dots,v_p^*\in V^*$ be covectors. Then \begin{align*} \langle\ext_{v_1}(v_2\wedge\cdots\wedge v_p),v_1^*\wedge\cdots\wedge v_p^*\rangle &=\langle v_1\wedge\cdots\wedge v_p,v_1^*\wedge\cdots\wedge v_p^*\rangle\\ &=\langle v_1\otimes\cdots\otimes v_p,v_1^*\wedge\cdots\wedge v_p^*\rangle\psi(p), &\text{by Exercise 16,}\\ &=\langle \tsr_{v_1}(v_2\otimes\cdots\otimes v_p),v_1^*\wedge\cdots\wedge v_p^*\rangle\psi(p)\\ &=\langle v_2\otimes\cdots\otimes v_p,\int_{v_1}(v_1^*\wedge\cdots\wedge v_p^*)\rangle\psi(p), &\text{by \eqref{eq},}\\ &=\langle v_2\wedge\cdots\wedge v_p,\int_{v_1}(v_1^*\wedge\cdots\wedge v_p^*)\rangle,&\text{by Exercise 16.}\square \end{align*}

3. The adjoints of $\tsr_{v^\flat}$ and $\ext_{v^\flat}$

We are going to consider now a specialized case of the previous adjunctions that is relevant for differential geometry. Now, we are going to fix a non-degenerate symmetric bilinear form $g:V\times V\to k$. This gives us an isomorphism \begin{align*} L_g=R_g=(-)^\flat:V&\longrightarrow V^*\\ v&\longmapsto v^\flat=g(v,-)=g(-,v). \end{align*} Denote $(-)^\sharp: V^*\to V$ to the inverse of $(-)^\flat$.

Since the tensor algebra is a covariant functor, the maps $(-)^\flat: V \rightleftarrows V^* :(-)^\sharp$ induce inverse algebra isomorphisms $$ (-)^\flat: TV \rightleftarrows TV^* :(-)^\sharp. $$ The $k$-duality \eqref{TV_dlty} induces a $k$-duality $$ \tag{9}\label{eq:dlty_TV_dual} \langle -,-\rangle:TV^*\times TV^*\to k. $$ Namely, the one that makes the following diagram commute: $$\require{AMScd} \begin{CD} TV^*\times TV^* @>{\langle -,-\rangle}>> k \\ @V{((-)^\sharp,1_{TV^*})}VV @| \\ TV\times TV^* @>>{\langle -,-\rangle}> k \end{CD}$$

Thus, by Lemma 13, the adjunction $ (\tsr_v, \int_v, \langle -,-\rangle_{TV,TV^*}) $ induces an adjunction $$ ((-)^\sharp\circ\tsr_v\circ(-)^\flat, \int_v, \langle -,-\rangle_{TV^*,TV^*}). $$ We claim that $(-)^\sharp\circ\tsr_v\circ(-)^\flat=\tsr_{v^\flat}$. Proving this amounts to show commutativity of the diagram $$ \tag{10}\label{eq:diag_tsr_flat} \require{AMScd} \begin{CD} TV @>{\tsr_v}>> TV \\ @V{(-)^\flat}VV @VV{(-)^\flat}V \\ TV^* @>>{\tsr_{v^\flat}}> TV^* \end{CD}$$ By the universal property of the tensor algebra, it suffices to show commutativity for elements of $V\subset TV$. But this is trivial since $(-)^\flat:TV\to TV^*$ is an algebra homomorphism and so $(v\otimes w)^\flat=v^\flat\otimes w^\flat$. This shows that $ (\tsr_{v^\flat}, \int_v, \langle -,-\rangle_{TV^*,TV^*}) $ is an adjunction.

Imitating what we just did and mutatis mutandis, from the duality \eqref{dlty_ext_pow} we can obtain a $k$-duality $$ \tag{11}\label{eq:dlty_lambda_V_dual} \langle -,-\rangle :\Lambda V^*\times\Lambda V^*\to k, $$ leveraging now on the fact that the exterior algebra assignment is a functor.

Claim 18. The duality \eqref{eq:dlty_lambda_V_dual} is a subduality of \eqref{eq:dlty_TV_dual}.

Proof. We want to show that the upper triangle in the following diagram commutes:

For this, it suffices to show that the rest of the cells of the diagram commute. The left and right triangles commute by definition. The outer square commutes by naturality of the inclusion $\Lambda W\hookrightarrow TW$ (which amounts to naturality of the inclusions $\Lambda^pW\hookrightarrow T^pW$). Lastly, the lower triangle commutes because $\Lambda V\times\Lambda V^*\hookrightarrow TV\times TV^*$ is a subpairing, (see comments after \eqref{dlty_ext_pow}). $\square$

We then obtain an adjunction $ (\ext_{v^\flat}, \int_v, \langle -,-\rangle_{\Lambda V^*,\Lambda V^*}) $ from the adjunction $ (\ext_{v}, \int_v, \langle -,-\rangle_{\Lambda V,\Lambda V^*}) $. The analogous diagram for \eqref{eq:diag_tsr_flat} would now be $$ \require{AMScd} \begin{CD} \Lambda V @>{\ext_v}>> \Lambda V \\ @V{(-)^\flat}VV @VV{(-)^\flat}V \\ \Lambda V^* @>>{\ext_{v^\flat}}> \Lambda V^* \end{CD}$$ which can be seen to commute using the universal property of the exterior algebra: it suffices to verify commutativity for elements of $V\subset\Lambda V$. But we have $(v\wedge w)^\flat=v^\flat\wedge w^\flat$, for $(-)^\flat:\Lambda V \to \Lambda V^*$ is an algebra homomorphism.

As a conclusion to this post, we are going to describe in a more explicit way what the pairings $\langle -,-\rangle_{TV^*,TV^*}$ and $\langle -,-\rangle_{\Lambda V^*,\Lambda V^*}$ look like in terms of $g$, in a more explicit way.

If we apply Lemma 15 to $g$, we obtain a non-degenerate symmetric bilinear form \begin{align*} T^pg:T^pV\times T^pV&\to k\\ (v_1\otimes\dots\otimes v_n,w_1\otimes\cdots\otimes w_n) &\mapsto g(v_1,w_1)\cdots g(v_n,w_n). \end{align*} By Lemma 14, these assemble into a non-degenerate bilinear form $$ Tg: TV\times TV\to k. $$ On the other hand, we have an induced non-degenerate symmetric bilinear form $g^\sharp:V^*\times V^*\to k$ given by $g^\sharp(\omega,\eta)=g(\omega^\sharp,\eta^\sharp)$, so we get another non-degenerate symmetric bilinear form $$ Tg^\sharp: TV^*\times TV^*\to k. $$

Claim 19. It holds $\langle -,-\rangle_{TV^*,TV^*}=Tg^\sharp(-,-)$.

Proof. We want to show that the outer triangle of the following diagram commutes:

For this, in turn, it suffices to show that the three internal triangles commute. The upper triangle commutes by definition and the left triangle obviously commutes. We will verify then that the lower triangle commutes.

We must see that this lower triangle commutes for $k$-vectors paired with $k$-covectors in $TV\times TV^*$. First consider the case $k=1$. Let $v\in V$ and $\omega\in V^*$. Then \begin{align*} g(v,\omega^\sharp)&=g(-,\omega^\sharp)(v)\\ &=(\omega^{\sharp})^{\flat}(v)\\ &=\omega(v)\\ &=\langle v,\omega\rangle. \end{align*} Now let $v_1,\dots,v_p\in V$ and $\omega_1,\dots,\omega_p\in V^*$. Then \begin{align*} Tg(v_1\otimes\cdots\otimes v_p, (\omega_1\otimes\cdots\otimes\omega_p)^\sharp) &=Tg(v_1\otimes\cdots\otimes v_p, \omega_1^\sharp\otimes\cdots\otimes\omega_p^\sharp)\\ &=g(v_1,\omega_1^\sharp)\cdots g(v_p,\omega_p^\sharp)\\ &=\langle v_1,\omega_1\rangle\cdots\langle v_p,\omega_p\rangle\\ &=\langle v_1\otimes\cdots\otimes v_p, \omega_1\otimes\cdots\otimes\omega_p\rangle.&\qquad\square \end{align*}

In particular, this shows that $\langle -,-\rangle_{TV^*,TV^*}$ is symmetric. Thus, since $ (\tsr_{v^\flat}, \int_v, \langle -,-\rangle_{TV^*,TV^*}) $ is an adjunction, so is $ (\int_v, \tsr_{v^\flat}, \langle -,-\rangle_{TV^*,TV^*}) $.

Denote $\Lambda g=Tg|_{\Lambda V}:\Lambda V\times\Lambda V\to k$. The following is proven in a similar way as Exercise 16.

Exercise 20. Given vectors $v_1,w_1,\dots,v_n,w_n\in V$, show \begin{align*} T^ng( v_1\wedge\cdots\wedge v_n,w_1\otimes\cdots\otimes w_n) &=\frac{\psi(n)}{n!}\det(g( v_i,w_j))=T^ng( v_1\otimes\cdots\otimes v_n,w_1\wedge\cdots\wedge w_n),\\ T^ng( v_1\wedge\cdots\wedge v_n,w_1\wedge\cdots\wedge w_n) &=\frac{\psi(n)^2}{n!}\det(g( v_i,w_j)). \end{align*}

Usage of claims 18 and 19 shows then that $\langle -,-\rangle_{\Lambda V^*,\Lambda V^*} =\Lambda g^\sharp(-,-)$. In particular, $\langle -,-\rangle_{\Lambda V^*,\Lambda V^*}$ is symmetric, and, thus, since $ (\ext_{v^\flat}, \int_v, \langle -,-\rangle_{\Lambda V^*,\Lambda V^*}) $ is an adjunction, so is $ (\int_v, \ext_{v^\flat}, \langle -,-\rangle_{\Lambda V^*,\Lambda V^*}) $.

On the other hand, by the last exercise, for $\omega_1,\eta_1,\dots,\omega_n,\eta_n\in V^*$, we have $$ \langle\omega_1\wedge\cdots\wedge\omega_n, \eta_1\wedge\cdots\wedge\eta_n\rangle = \frac{\psi(n)^2}{n!}\det(g(\omega_i^\sharp,\eta_j^\sharp)). $$

Note that if we redefined the pairing $\langle-,-\rangle_{\Lambda V^*,\Lambda V^*}$ to be $\frac{n!}{\psi(n)^2}$ times the old pairing then we would get rid of the normalization constant in the last equality and $(\ext_{v^\flat},\int_v,\langle-,-\rangle_{\Lambda V^*,\Lambda V^*})$ would still be an adjunction. In fact, there are authors that begin with a finite-dimensional $k$-vector space $V$ equipped with a non-degenerate symmetric bilinear form $g$ on $V$ and then define the pairing $\langle-,-\rangle_{\Lambda V^*,\Lambda V^*}$ explicitly in this way. However, the adjunction is now not so obvious.