# DDG Week 4

## k-forms

Week 3: Exterior algebra (k-vectors) Week 4: Exterior calculus (k-forms)

Exterior Calculus, "how do lengths, areas, volumes, change over curved surfaces?

Measurement devices have the same dimensionality as the thing we're measuring.

Study exterior calculus in flat spaces ($\mathbb{R}^n$) May seem redundant, but will help when space starts curving.

### Vectors & Covectors

covectors do measurements vectors get measured

Wedging covectors get k-forms, which are duals to k-vectors.

Similar to row/column vectors. A row vector is a linear map which maps the column vector to a real value.

let $\alpha$ be a unit covector, $u$ be a vector of any magnitude. $\alpha(u)$ tells us how long $u$ is in the direction $\alpha$

Defn:

Let $V$ be a ceal vector space. Its dual space $V^*$ is the collection of all linear functions $\alpha : V \mapsto \mathbb{R}$ together with the operations of addition.

$(\alpha + \beta)(u) := \alpha(u) + \beta(u)$

and scalar multiplication

$(c \alpha)(u) := c (\alpha(u))$

$\forall \alpha,\beta \in V^*, u \in V, c \in \mathbb{R}$

Defn:

An element of a dual vector space is called a dual vector or covector

(unrelated to the Hodge Star/Dual)

e.g $V = C^0$ and $\delta(f) := f(0) \in V^*$ (Dirac delta)

Let $[x,y,z] [r,g,b]$ be column vectors, then the transpose converts between row/column. Likewise, let $u,v \in V$ be vectors, then $u^\flat, v^\flat \in V^*$. let $\alpha, \beta \in V^*$ be covectors, then $\alpha^\sharp, \beta^\sharp \in V$.

let the inner product be $a^T M b$ instead of the normal inner product. $u^\flat(v) = u^T M v$ and $\alpha(\beta^\sharp) = \alpha M^{-1} \beta^T$ Equivalently, $u^\flat(\cdot) = $ and $\alpha(\cdot) = <\alpha^\sharp, \cdot>$

k-forms will be multilinear maps from a k-vector.

### Measurement of 2-vectors

The "shadow" of a parallelogram onto another parallelogram. Given $(u,v)$, pick orthonormal basis for plane $\alpha, \beta$.

1. project onto plane
• $u \mapsto (\alpha(u), \beta(u))$
• $v \mapsto (\alpha(v), \beta(v))$
1. calculate area via cross product
• $\alpha(u)\beta(v) - \alpha(v)\beta(u)$

Can apply same expression when $\alpha, \beta$ not orthonormal.

$(\alpha \wedge \beta)(u,v) := \alpha(u)\beta(v) - \alpha(v)\beta(u)$

This is the definiton of the application of a 2-form to two vectors. Note: result scales with "area" of $\alpha \wedge \beta$

Note that the application of the 2-form is antisymettric

$(\alpha \wedge \beta)(u,v) = - (\alpha \wedge \beta)(v,u)$

Geometrically, this shows that the orientation of the face is different.

The arguments to the wedge are also antisymmetric.

$(\alpha \wedge \beta)(u,v) = -(\beta \wedge \alpha)(u,v)$

this has the same meaning, as the 2-form's orientation is flipped.

### Measurement of 3-vectors

in $\mathbb{R}^3$, all 3-vectors have the same "direction". Can only measure magnitude.

Let $u,v,w$ be the edges of a parallelopiped. Let $(\alpha, \beta, \gamma)$ be any orthonormal basis.

1. project $u,v,w$ onto the basis
• $u \mapsto (\alpha(u), \beta(u), \gamma(u))$
• $v \mapsto (\alpha(v), \beta(v), \gamma(v))$
• $w \mapsto (\alpha(w), \beta(w), \gamma(w))$
1. apply standard formula for volume (det)
$\alpha(u)\beta(v)\gamma(w) + \alpha(v)\beta(w)\gamma(u) + \alpha(w)\beta(u)\gamma(v) \\ - \alpha(u)\beta(w)\gamma(v) - \alpha(w)\beta(v)\gamma(u) - \alpha(v)\beta(u)\gamma(w)$

Generally, a k-form is a fully antisymmetric, multilinear measurement of a vector.

Typically, think of a k-form as a map from a k-vector to a scalar.

$\alpha : V \times \cdots V \rightarrow \mathbb{R}$
$(\alpha_1 \wedge \cdots \wedge \alpha_k)(u_1, \dots, u_k) := \det((\alpha_1(u_1), ..., \alpha_k(u_k)), \dots, (\alpha_k(u_1), ..., \alpha_k(u_k)))$

For historical reasons, we do not write

$(\alpha \wedge \beta) (u \wedge v)$

and instead use the notation we have been using above.

note: a 0-form is a scalar

### k-forms in coordinates

For vectors in a basis $e_1, \dots, e_n$, write

$v = v^1 e_1 + \cdots + v^n e_n$

The scalar values $v^i$ are the coordinates of v.

For covectors in a so-called dual basis $e^1, \dots, e^n$, write

$\alpha = \alpha_1 e^1 + \cdots + \alpha_n e^n$

Where

$e^i(e_j) = \delta_{ij}$

where $\delta_{ij}$ is the Kronecker delta.

ex.

$v = 2 e_1 + 2 e_2$ $\alpha = -2 e^1 + 3 e^2$

$\alpha(v) = (-2e^1 + 3e^2)(2e_1 + 2e_2) \\ \quad = -2e^1(2e_1 + 2e_2) + 3e^2(2e_1 + 2e_2) \\ \quad = -4 e^1(e_1) - 4e^1(e_2) + 6e^2(e_1) + 6e^2(e_2) \\ \quad = -4 + 6 = 2$

ex. 2-form

$u = 2e_1 + 2e_2 \quad \alpha = e^1 + 3e^2 \\ v = -2e_1 + 2e_2 \quad \beta = 2e^1 + e^2$

Then,

$(\alpha \wedge \beta)(u,v) = \alpha(u)\beta(v) - \alpha(v)\beta(u) \\ \alpha(u) = 2 + 6 = 8\\ \alpha(v) = -2 + 6 = 4\\ \beta(u) = 4 + 2 = 6\\ \beta(v) = -4 + 2 = -2$

Thus,

$(\alpha \wedge \beta)(u,v) = 8\cdot(-2) - (4)\cdot6 = -40$

Recall $\sharp \text{ and } \flat$ If $\alpha^\sharp = u ~\land~ u^\flat = \alpha$ then

$\alpha = a_1 e^1 + \cdots + a_ne^n \stackrel{\sharp}{\implies} u = u^1e_1 + \cdots + u^ne_n$ $\alpha = a_1 e^1 + \cdots + a_ne^n \stackrel{\flat}{\impliedby} u = u^1e_1 + \cdots + u^ne_n$

This is sometimes called the musical isomorphism