Cadabra SW: Applying to some exercises in Nakahara

Tests on differential forms:


Test 1: If $\omega$ is a differential form of odd dimension - say 3, then $\omega  \wedge \omega = 0$. This is the attempt to let Cadbra solve this wedge form.


Nakahara eqn 5.67a

Cadabra code:

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{a,b,c,l,m,n}::Indices.

{e^{a}, \omega^{a}_{b}}::DifferentialForm(degree=3);

$\displaystyle{}\text{Attached property DifferentialForm to }\left[e^{a}, \omega^{a}\,_{b}\right].$

eq1 := \omega^{a}_{b} ^ \omega^{a}_{b};

0

Nice! Solves this. Test 2: Now we want to show explicitly - that is using numeric indices for $q,r$ the following expression:

$\eta \wedge \nu = (-1)^{qr} \nu \wedge \eta$.

Cadabra code:

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def post_process(ex):

sort_product(ex)

canonicalise(ex)

collect_terms(ex)

{ \eta^{a}_{b}}::DifferentialForm(degree=3);

{ \nu^{a}_{b}}::DifferentialForm(degree=5);

$\displaystyle{}\text{Attached property DifferentialForm to }\eta^{a}\,_{b}.$

$\displaystyle{}\text{Attached property DifferentialForm to }\nu^{a}\,_{b}.$

eq2 := \eta^{a}_{b} ^ \nu^{a}_{b}; eq3 := \nu^{a}_{b} ^ \eta^{a}_{b};

$\displaystyle{}\eta^{a}\,_{b}\wedge \nu^{a}\,_{b}$

\eta^{a}_{b} ^ \nu^{a}_{b}

$\displaystyle{}-\eta^{a}\,_{b}\wedge \nu^{a}\,_{b}$

-\eta^{a}_{b} ^ \nu^{a}_{b}

Cadabra code:

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eq2 + eq3;

combine(_);

$\displaystyle{}\eta^{a}\,_{b}\wedge \nu^{a}\,_{b}-\eta^{a}\,_{b}\wedge \nu^{a}\,_{b}$
\eta^{a}_{b} ^ \nu^{a}_{b}-\eta^{a}_{b} ^ \nu^{a}_{b}

$\displaystyle{}0$

0





Exercise 5.15: Let $\xi \in \Omega^{q}(M)$ and $\omega \in \Omega^{r}(M)$.

Show that $d(\xi \wedge \omega) = d\xi \wedge \omega + (-1)^{qr} \xi \wedge d\omega$.

For simplicity, we shall set $q=3$ and $r=5$ - thus inducing a negative in the expression.
Using the following link is nice: https://cadabra.science/notebooks/exterior.html

Cadabra code:

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\xi::DifferentialForm(degree=3);

\omega::DifferentialForm(degree=5);

$\displaystyle{}\text{Attached property DifferentialForm to }\xi.$
$\displaystyle{}\text{Attached property DifferentialForm to }\omega.$


Add definition of exterior derivative.

Cadabra code:

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d{#}::ExteriorDerivative;

d{#}::LaTeXForm("{\rm d}").


$\displaystyle{}\text{Attached property ExteriorDerivative to }d{\#}.$
ext1 := d{ \xi ^ \omega };
$\displaystyle{}{\rm d}\left(\xi\wedge \omega\right)$

Cadabra code:

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d(\xi ^ \omega)
product_rule(_);


$\displaystyle{}{\rm d}{\xi}\wedge \omega-\xi\wedge {\rm d}{\omega}$
d(\xi) ^ \omega-\xi ^ d(\omega)


This demonstrates equation $5.69$ in Nakahra for even and odd indices.