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I'm an undergraduate student and currently I'm approaching tensorial calculus. I was wondering: is there some geometric meaning to the operation of rising/lowering indices (and then if there was any geometric difference between vectors and covectors), or are they only mere formal operations? In case, why don't we simply use only vectors in the definition of tensors?

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    $\begingroup$ Have you learned what the dual vector space is? That's crucial to understanding what tensors are. In short, if a vector space has an inner product (e.g., $\mathbb{R}^n$ with the standard dot product), then there is a natural isomorphism from the vector space to its dual. Raising or lowering an index is effectively transforming a tensor using this isomorphism. $\endgroup$
    – Deane
    Commented Aug 6, 2023 at 15:44
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    $\begingroup$ @Deane Yes, I know dual vector spaces, but I did't catch its geometric meaning, if it has one. Moreover, what's the need of use both vectors and their duals? $\endgroup$
    – Amenable
    Commented Aug 6, 2023 at 15:53

3 Answers 3

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$ \newcommand\R{\mathbb R} \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\doub{\mathfrak} \newcommand\Blades[1]{\mathop{\textstyle\bigwedge^{\mkern-2mu(#1)}}} \newcommand\MVects[1]{\mathop{\textstyle\bigwedge^{\mkern-1mu#1}}} \newcommand\lintr{\mathbin{\lrcorner}} $The other answers I see miss the geometry. Let $V$ be a finite-dimensional real vector space, though this discussion is largely independent of the field.

A vector $0 \ne v \in V$ is, in a sense, an oriented, weighted line. It is oriented in the sense that we have both $v$ and $-v$, it is weighted in the sense that $w + av$ for any other $w$ is "more like $v$" as $a \to \infty$, and it is a line in the sense that $\mathrm{span}\{v\}$ is a line. (Making the phrase "more like $v$" precise requires something like a metric.)

An covector $0 \ne \alpha \in V^*$ is an oriented, weighted hyperplane, i.e. a subspace of dimension $\dim V - 1$. It is oriented and weighted in the same sense that vectors are, and it is a hyperplane in the sense that $\ker\alpha = \{v \in V \;:\; \alpha(v) = 0\}$ is a hyperplane.

A metric is, for our purposes, a nondegenerate symmetric bilinear form $g : V \times V \to \R$. We get a map $V \to V^*$, denoted $v \mapsto v^\flat$, via $$ v^\flat(w) = g(v, w). $$ Because $g$ is nondegenerate, this map is actually an isomorphism $V \cong V^*$ and its inverse is denoted by $\alpha \mapsto \alpha^\sharp$. These flat and sharp operators are what lower and raise indices: $$ (v^\flat)_i = g_{ij}v^j,\quad (\alpha^\sharp)^i = g^{ij}\alpha_j. $$

But an isomorphism $V \cong V^*$ is a one-to-one correspondence between lines and hyperplanes. What is this correspondence? It is orthogonalization. Let $L, H$ be the sets of lines, hyperplanes: $$ L = \{\mathrm{span}\{v\} \;:\; v \in V\},\quad H = \{\ker\alpha \;:\; \alpha \in V^*\}. $$ Then the isomorphism $V \cong V^*$ becomes a bijection $L \cong H$ via $$ \mathrm{span}\{v\} \mapsto \ker v^\flat. $$ This is well defined because $\ker(av)^\flat = \ker(av^\flat) = \ker v^\flat$ for any nonzero scalar $a$. But now suppose $w \in \ker v^\flat$; trivially $$ g(v, w) = v^\flat(w) = 0 $$ so every element of $\ker v^\flat$ is orthogonal to $v$ and indeed $\ker v^\flat = (\mathrm{span}\{v\})^\perp$.

In summary:

  • Lowering an index is the operation $v \mapsto v^\flat$ which takes a line (represented by a vector $v$) to its orthogonal hyperplane $(\mathrm{span}\{v\})^\perp = \ker v^\flat$.
  • Raising an index is the operation $\alpha \mapsto \alpha^\sharp$ which takes a hyperplane (as represented by a covector $\alpha$) to its orthogonal line $(\ker\alpha)^\perp = \mathrm{span}\{\alpha^\sharp\}$.

What does $V^*$ accomplish?

"Why do we need covectors and can't just use only vectors?" And the simple answer is that when you have a metric, or rather an isomorphism $V \cong V^*$, you don't need covectors at all.

But we aren't interested in just one metric; we are interested in many, many different metrics which need not be equivalent to each other in any reasonable sense. We are also interested in bilinear forms which aren't symmetric, or which may be degenerate. The theory of the pair $(V^*, V)$ is a neat way of reasoning about the theory of not just bilinear forms, but of multilinear maps and how they interact. Actually, it often all comes down to the double space $\doub W := V^*\oplus V$. You could consider it an organizational principle.

In the interest of being brief and accessible though, I will not get into details. What I will do though is sketch the following idea, which is only really ellucidated by considering the pair $(V^* ,V)$:

The Hodge star of a metric $g$ is the composition of the following two ideas:

  1. Subspaces can be built up by joining smaller subspaces, or built down by meeting (i.e. intersecting) larger subspaces. In particular, we build up to any subspace using lines, and build down to any subspace using hyperplanes.
  2. The views of Building up and building down are precisely related by orthogonalization: the orthogonal complement of the join of two subspaces is the meet of their orthogonal complements, and vice versa: $$ (A\oplus B)^\perp = A^\perp\cap B^\perp,\quad (A\cap B)^\perp = A^\perp\oplus B^\perp. $$ (Of course, $A\cap B = \{ 0 \}$)

The Hodge Star

The exterior algebra $\Ext V$ of a vector space $V$ can be characterized as the unique-up-to-isomorphism associative algebra generated by $V$ and the identity $1$ such that $$ v_1\wedge\dotsb\wedge v_k = 0 \iff \text{those vectors are linearly dependent}. $$ This allows us to associate a product of $k$ vectors $X \in \Blades kV$ (hence forth called a $k$-blade) with a $k$-subspace of $V$ $$ [X] := \{v \in V \;:\; v\wedge X =0 \} $$ in the sense that we have a bijection $$ \{\mathbb RX \;:\; 0\ne X \in \Blades kV\} \longleftrightarrow \{S \subseteq V \;:\; S\text{ is a subspace and }\dim S = k\} $$ given by $\mathbb RX \mapsto [X]$. Under this identification, the wedge product is an operation on subspaces $S, T$ which yields zero when $S\cap T \ne \{ 0 \}$ and otherwise yields the "join" $S\oplus T$. I should note that $\Blades kV$ is not a vector space; its span $\MVects kV$ is callled the space of $k$-vectors, and the algebra is graded on $k$-vectors: $$ \Ext V = \mathbb R\oplus V\oplus\MVects 2V\oplus\MVects 3V\oplus\dotsb \oplus\MVects nV $$ where $\MVects 0V = \mathbb R$, $\MVects 1V = V$, and $n = \dim V$. Note that the isomorphism between blades and subspaces crucially implies that $\MVects nV$ is a one dimensional space because there is only one $n$-dimensional space: $V$ itself. For this reason $\MVects nV$ is called the space of pseudoscalars.

Though our hand is forced along the way, it is by no small feat that we can arrive at the dual view of $\Ext V^*$: as the elements of $V^*$ are hyperplanes, the wedge product on this space is an operation on subspaces $S, T$ which yields zero when $S, T$ are contained in a common hyperplane and otherwise yields the meet $S\cap T$. In brief, dual (in fact adjoint) to how the wedge product is grade-raising, we can define a grade-lowering operation $\lintr : \Ext V\times\Ext V^* \to \Ext V^*$ called the left interior product; the map between blades $X^*$ and subspaces is then $$ [X^*] = \{v \in V \;:\; v\lintr X^* = 0\}. $$ In this way we have bijections between the $(n-k)$-blades of $V^*$ and the $k$-blades of $V$ $$\begin{aligned} \{\R X^* \;:\; 0 \ne X^* \in \Blades{n-k}V^*\} &\longleftrightarrow \{S \;:\; S\text{ is a subspace of $V$ and }\dim S = k\} \\ &\longleftrightarrow \{\R X \;:\; 0 \ne X \in \Blades kV\}. \end{aligned}$$ These bijections build naturally into a (vector space) isomorphism $P : \Ext V^*\cong\Ext V$ called the Poincaré isomorphism. This is part (1) above.

However, a metric $g : V\times V \to \R$ is an isomorphism $\flat : V \cong V^*$, and it is quite simple to extend this to an algebra isomorphism $\Ext V \cong \Ext V^*$ via $$ (v_1\wedge\dotsb\wedge v_k)^\flat = v_1^\flat\wedge\dotsb\wedge v_k^\flat. $$ This is just an extension of the orthogonalization idea from earlier: it maps $k$-vector representing a $k$-space to its orthogonal $(n-k)$-space as represented by a $k$-covector. This is part (2) above.

The Hodge star $\star : \Ext V^* \to \Ext V^*$ of differential forms is precisely $$ \star = \flat\circ P $$ mapping $k$-covectors to $(n-k)$-covectors.

In this way may construct and reason about all Hodge star maps by reasoning about the single map $P$ and relegating any metric-specific concerns to its interaction with $\flat$. This is one way in which considering the pair $(V^*, V)$ gives a nice organization and separation of concerns.

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    $\begingroup$ Should be the accepted answer. $\endgroup$
    – blargoner
    Commented Aug 7, 2023 at 1:31
  • $\begingroup$ @Amenable I realized I didn't answer the "why covectors" part of your question and have added a section to my answer. $\endgroup$ Commented Aug 7, 2023 at 19:23
  • $\begingroup$ @NicholasTodoroff Thank you for your great answer, now it's crystal clear!! $\endgroup$
    – Amenable
    Commented Aug 9, 2023 at 7:33
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There is a difference between tangent vectors and covectors. The fast way to explain it is that both vectors and covectors are a differential object, namely it suffices to have a $C^1$-manifold to define them, while it is possible to rise and lower index only adding further structure, namely endowing the manifold with a pseudo-Riemannian metric.

As @Deane remarked in the comment, the main point is understanding that the cotangent space is the dual of the tangent, and that there is no natural isomorphism between a vector space and its dual.

However, if the vector space has a non-degenerate inner product $\langle\cdot,\cdot\rangle$, one can identify the two space by associating to a vector $v$ the covector $\langle v,\cdot\rangle$.

A (pseudo-)metric is a non-degenerate inner product on the tangent space, hence it allows to build this identification.

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  • $\begingroup$ Thanks to you and to Deane, now its more clear, but I still need to review the theory of dual spaces. $\endgroup$
    – Amenable
    Commented Aug 6, 2023 at 16:16
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Let's first discuss the linear algebra. Let $V$ be a real vector space, and define $$V^*:=\{\text{linear maps from } V \text{ to } \mathbb{R}\}.$$ Call elements in $V^*$ covectors. Observe that $V^*$ is again a vector space since function addition (scalar multiplication) is defined as pointwise addition (multiplication) of real numbers. Also, the zero function serves as the zero element in $V^*$. If $V$ carries a basis $\{e_i\}$, there is canonically an induced basis $\{e^i\}$ of $V^*$ defined by the formula $$e^i(e_j)=\delta^i_j:=\begin{cases}1&i=j\\ 0&i\neq j.\end{cases}$$

Now suppose $V$ carries an inner product, that is a certain non-degenerate bilinear form $$\langle -,-\rangle:V\times V\to \mathbb{R}.$$ For every vector $v\in V$, you may "freeze" one slot at $v$ to get a linear map $\langle v,-\rangle:V\to \mathbb{R}$. That is, $\langle v,-\rangle\in V^*$. Thus, we may instead view an inner product as the liner map

\begin{align}L:V&\to V^*\\ v&\mapsto \langle v,-\rangle. \end{align}

The non-degeneracy condition in the inner product implies that this map is an isomorphism, that is an identification between $V$ and $V^*$. Different choices of inner products give different choices of identifications between $V$ and $V^*$. An inner product on $V$ also induces one on $V^*$ in a way I'll discuss more in detail soon.

Now let's discuss the differential geometry. When you have local coordinates $\{x^i\}$, you get a (local) basis $\{\partial_i\}$ on the tangent spaces. By the linear algebra discussed before, you get a (local) basis $\{dx^i\}$ on the dual spaces of the tangent spaces, hereby called the cotangent spaces. Suppose you have a smoothly varying choice of inner product $g$, and define wrt the local coordinates $\{x^i\}$ the functions $$g_{ij}:=g(\partial_i,\partial_j).$$ Together, $g_{ij}$ defines (at each point in your local chart) a (smoothly varying) symmetric, invertible matrix. Define $g^{ij}:=(g^{-1})_{ij}$, the $ij$th entry of the inverse matrix. This defines the induces the inner product on the dual space, $$\langle dx^i,dx^j\rangle=g^{ij}.$$

For example, suppose we have a $(0,2)$-tensor $T$ where we write locally $$T=T_{ij}dx^i \otimes dx^j.$$ The operation of raising an index is to identify $T$ with the $(1,1)$-tensor $$T^k_j\partial_k \otimes dx^j:=g^{ki}T_{ij}\partial_k \otimes dx^j.$$ At each point $p\in M$, you're using $L$ ($L^{-1}$) if you're lowering (raising) indices to change a vector $v\in T_pM$ (covector $\omega \in T_p^*M$) to the covector $Lv=\langle v,- \rangle\in T_p^*M$ (vector $L^{-1}\omega\in T_pM$). In our example, when we raised the index, we used $L^{-1}$ to identify $$v=T_{ij}dx^i\in T_p^*M$$ with $$L^{-1}v=g^{ki}T_{ij}\partial_k\in T_pM.$$ I'll leave it to you to expand this out in terms of the relevant bases to see that this is indeed applying $L^{-1}$. You should also go through an example of lowering an index to see that it really is applying $L$.

As you suggest, if we have a metric we can keep on raising (lowering) indices of a $(p,q)$-tensor to obtain a $(p+q,0)$-version (a $(0,p+q)$) of the same tensor. But from what I've seen, this is mostly an aesthetic change - it's not like you'll understand the object better if you raise (lower) all the indices. For example, one person may write Bocher's formula as $$\frac{1}{2}\Delta |\nabla u|^2=\text{Ric}(\nabla u,\nabla u)+\langle \nabla u,\nabla \Delta u \rangle+|\nabla^2 u|^2,$$ whereas another person may write the same formula as $$\frac{1}{2}\Delta |\nabla u|^2=\text{Ric}(du, du)+\langle du, d\Delta u\rangle+|D du|^2.$$ Both versions of Bochner's formula convey the same amount of information. The preference of which version you write is a matter of aesthetics - do you like gradients or exterior derivatives more?

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