Cadabra Q&A - Recent questions and answers

Answered: Tutorial on Differential Forms: I would expect A1 ^ A2 + A2 ^ A1 = 0

Sat, 02 May 2026 18:22:49 +0000

In the tutorial, A1 is a one-form (degree=1) and A2 is a two-form (degree=2). For those, the wedge product is symmetric. You are probably thinking of the wedge product of two one-forms.

Answered: Problem with SympyBridge

Mon, 13 Apr 2026 14:35:52 +0000

It chokes on the backslash in the \Dg, because Python does not like backslashes in variable names. All "normal" LaTeX symbols get (e.g. \alpha) get their backslash stripped and then re-instated afterwards, but if you use custom symbols that will not work.

Since you are already using a LaTeXForm to make the output look nice, I would just use Dg instead of \Dg everywhere. That will work.

Answered: Cadabra on Windows 10 (2026) - integrate into Jupyter Notebook

Fri, 03 Apr 2026 13:05:06 +0000

In general you're better off using the "new" Jupyter kernel for Cadabra which is built by default. On Linux, this is a lot easier than using the old Xeus-based kernel in a conda setup. I will soon remove that support from the source tree altogether, as conda is more and more problematic and there is nobody maintaining those packages anymore.

On Windows, you really are much much better off by running the jupyter notebook server inside WSL, so you can rely on the Linux packages. Then connect to that via a browser on the Windows side.

Answered: Trouble running standard cnb file

Thu, 02 Apr 2026 10:45:14 +0000

As for your other question: in

t1 = action[1][0][0][2]

we start from the equation action. This is of the form $$ S = \int \left(\cdots\right) {\rm d}x. $$ The action[1] takes the first child (counting from zero), which is the right-hand side. Could also have been written as rhs(action). Then the zeroth child of that is the integrand, the zeroth child of the integrand is the first term, and the 2nd child of that is the thing that multiplies $\sqrt{-g} \delta g^{\mu]$ (the 2nd factor in the product).

As for the second line,

eom:= 2\kappa @(t1) = 0;

This defines a new expression eom, which is an equation. The @(t1) pulls the expression which is known to python as t1 (defined in the first line) into this new eom expression.

You may want to read at least the first chapter of the cadabra book, if the input format confuses you.

Answered: declare a mapping with previous expression

Mon, 02 Feb 2026 14:25:23 +0000

For some reason, the syntax @[something] or @(something) does not like underscores, so if you remove it from the variable name, the construction will work. For example:

M_mapping := M_{\mu} -> @(M_expr); <--- error

but

M_mapping := M_{\mu} -> @(Mexpr); <--- works

Answered: canonicalise gets stuck on swapped indices

Wed, 21 Jan 2026 19:42:01 +0000

Make sure to reset your kernel if you change properties like this. For me, the following works:

{\mu,\nu,\alpha,\beta}::Indices(position=fixed);
\nabla{#}::Derivative;
expr_test := \nabla_{\mu}{ A_{\alpha } } * \nabla^{\mu}{ A_{\beta} } 
           + \nabla^{\mu}{ A_{\alpha } } * \nabla_{\mu}{ A_{\beta} };
canonicalise(_);

and returns $2\nabla^{\mu} A_{\alpha} \nabla_{\mu} A_{\beta}$.

The logic is that with position=independent, the canonicalise algorithm is not allowed to raise/lower the $\mu$ indices in pairs. With position=fixed, this is allowed.

Answered: Rename specific dummy indices in expressions

Tue, 20 Jan 2026 15:18:31 +0000

Ok I ended up implementing something in line with the pseudo-code I was thinking:

from cdb.utils.indices import replace_index

def unify_variations_inplace(ex, symbol_name, target_indices):
    # 1. Helper to find the sum node (integrand)
    def find_sum_node(node):
        if node.name == r'\sum': return node
        for child in node.children():
            res = find_sum_node(child)
            if res: return res
        return None

    # Locate the node inside the original expression (ex)
    sum_node = find_sum_node(ex.top())
    if not sum_node: 
        return ex

    # 2. Build the unified integrand as a new Ex
    new_sum = Ex('0')
    for term_node in sum_node.children():
        t_ex = term_node.ex()
        matches = list(t_ex[symbol_name])

        if matches:
            current_indices = [str(idx.name) for idx in matches[0].indices()]
            for i in range(min(len(current_indices), len(target_indices))):
                old_idx, new_idx = current_indices[i], target_indices[i]
                if old_idx != new_idx:
                    t_ex = replace_index(t_ex, "", old_idx, new_idx)

        # We skip canonicalise here to prevent index shuffling
        new_sum += t_ex

    # 3. REINSERT the result into the original expression tree
    # This physically overwrites the sum inside the integral context
    sum_node.replace(new_sum)

    return ex

Then I use this like this:

dg{#}::LaTeXForm("\delta{g}").
action_test := \int{ W^{\mu}_{\nu \rho} g_{\mu \alpha} D^{\alpha} dg^{\rho \nu} + C_{\beta \mu} dg^{\beta \mu}}{x};
unify_variations_inplace(action_test, "dg", ['\psi', '\chi']);
factor_out(_, $dg^{\psi \chi}$);

The only shortcoming of this function is that as it is, it doesn't check if the target_indices are being already used inside the expression, but it shouldn't be too difficult to add

Issue with split_index and dummy index

Tue, 13 Jan 2026 11:06:33 +0000

I am new to Cadabra and have a couple of beginner questions that I haven’t been able to resolve on my own. I would greatly appreciate any help or clarification you can offer. I have the following Cadabra Code:

{M,N,P,Q,R#}::Indices(full, position=fixed).
{a,b,c,d,e#}::Indices(space1, position=fixed).
{a,b,c,d,e}::Integer(1..2).
{m,n,p,q,r#}::Indices(space2, position=fixed).

\partial{#}::PartialDerivative.
g_{M N}::Metric.
g^{M N}::InverseMetric.
g_{M?N?}::Symmetric.
g^{M?N?}::Symmetric.

Gtog:=\Gamma^{P?}_{M? N?}->(1/2)*g^{P? Q}(\partial_{N?}{g_{Q M?}}
+\partial_{M?}{g_{Q N?}}-\partial_{Q}{g_{M? N?}});

ex:=\Gamma^{2}_{1 2};
substitute(_, Gtog);
split_index(_, $M, a, m$);

todo:=\Gamma^{R}_{q m}\Gamma^{p}_{n R};
substitute(_, Gtog);

Question1: I tried using split index to split an index in my expression, but it doesn’t seem to do anything -- it simply returns the answer without performing the split. What might be the issue?

Question2: For the expression \Gamma^{R}{q m}\Gamma^{p}{n R} I am getting the same index Q repeated four times in the output. How can I prevent this and get the correct index contractions without the same dummy index appearing too many times?

Answered: absolute value in Cadabra expression

Mon, 22 Dec 2025 20:47:26 +0000

No algorithms in cadabra are currently aware of absolute values, so you can give it any name you want. But because \abs will not get converted to abs by the sympy bridge , it is probably best to simply use abs. In order to make it print as you want, use something like

abs{A??}::LaTeXForm("\left|", A??, "\right|");

The above will make the following work and print correctly:

ex := abs( -3 );
simplify(ex);

Answered: The mystery of meld

Thu, 04 Dec 2025 21:09:24 +0000

The first issue is now fixed on the devel branch.

Strange behavior young_project_product with zoom

Wed, 22 Oct 2025 16:56:49 +0000

Hello, I found a peculiar effect of Young's projection when applied to part of an expression after using zoom. For example:

{a,b,c,d,f,e, m,n,k,l,q,r,s,t,u,v,w,x,y,z,m#, n#}::Indices(values={0..3}, position=independent);

\nabla{#}::Derivative.
h_{m? n?}::Metric.
h^{m? n?}::InverseMetric.
h_{m? n?}::Symmetric.
h^{m? n?}::Symmetric.
F_{a? b?}::AntiSymmetric.
F^{a? b?}::AntiSymmetric.
\nabla_{a}{F_{b c}}::TableauSymmetry(shape={2,1}, indices={0,1,2});
\nabla_{a}{\nabla_{b}{p}}::Symmetric;

and I have some long expression:

expr := 1/2 F_{a b} F_{c d} R \nabla_{e}(p) \nabla_{f}(p) h^{a c} h^{b e} h^{d f}
-F_{a b} F_{c d} \nabla_{e}(\nabla_{f}(p)) \nabla_{m}(\nabla_{n}(p)) h^{a c} h^{b e} h^{d f} h^{m n} 
+ F_{a b} F_{c d} \nabla_{e}(\nabla_{f}(p)) \nabla_{m}(\nabla_{n}(p)) h^{a c} h^{b e} h^{d m} h^{f n}
-F_{a b} F_{c d} R_{e f} \nabla_{m}(p) \nabla_{n}(p) h^{a e} h^{b m} h^{c f} h^{d n}
-2F_{a b} \nabla_{c}(F_{d e}) \nabla_{f}(\nabla_{m}(p)) \nabla_{n}(p) h^{a f} h^{b n} h^{c d} h^{e m} 
+ F_{a b} F_{c d} \nabla_{e}(\nabla_{f}(p)) \nabla_{m}(\nabla_{n}(p)) h^{a e} h^{b m} h^{c f} h^{d n} 
+ F_{a b} \nabla_{c}(F_{d e}) \nabla_{f}(\nabla_{m}(p)) \nabla_{n}(p) h^{a c} h^{b d} h^{e f} h^{m n};

And I want to simplify terms with \nabla{a}{F{b c}}:

zoom(expr, $Q?? \nabla_{a}{F_{b c}}$);
expr2 = take_match($@[expr2]$, $Q?? \nabla_{a}{F_{b c}}$);
canonicalise(young_project_product(expr2));

For expr I get - 4/3 F_{a b} \nabla_{c}(F_{d e}) \nabla_{f}(\nabla_{m}(p)) \nabla_{n}(p) h^{a f} h^{b n} h^{c d} h^{e m} but for expr2 I get

- 8/9 F_{a b} \nabla_{c}(F_{d e}) \nabla_{f}(\nabla_{m}(p)) \nabla_{n}(p) h^{a f} h^{b n} h^{c d} h^{e m}

Answered: vary behaves in an odd way

Mon, 06 Oct 2025 20:34:44 +0000

The trivial cases always slip the net somehow... Now fixed on the devel branch, will go into 2.5.16. Thanks for reporting.

Answered: plot example on p. 51 of The Cadabra Book require range

Mon, 21 Jul 2025 17:05:48 +0000

For the range deduction to work you need to use the devel branch; this has not made it to the master branch yet but slipped into the docs inadvertedly.

You are correct about the samples parameter not being fed through correctly; I have opened https://github.com/kpeeters/cadabra2/issues/371 to keep track of this. Thanks.

Answered: \dot command not displayed correctly

Tue, 15 Jul 2025 11:10:52 +0000

Maybe I don't write it correctly but it's not working for me

can you show me the complete line ?

Answered: How to remove/unset properties

Sun, 13 Jul 2025 20:57:37 +0000

Thanks.

I clear command will be nice but probably not easy to implement.

Answered: Regex not working

Sun, 13 Jul 2025 20:15:49 +0000

If you have a fix please open an issue in github and attach a fix (or open a pull request).

(Regex is a feature I almost never use anymore, and was written before I started introducing rigid tests, so this does not surprise me at all).

Answered: How to collect similar terms?

Mon, 12 May 2025 16:30:12 +0000

Version 2.5.13 (currently in the devel branch on github) supports TeXmacs again. You need to put the config/init-cadabra2.scm file in the appropriate TeXmacs folder. Feedback is welcome.

Answered: execution problems

Tue, 08 Apr 2025 13:33:19 +0000

I don't think that we can make this work again, and I actually don't think this worked as you thought before. The python packaging logic is so terribly convoluted and broken that it is probably not a good idea to try to install Cadabra in CMAKE_INSTALL_PREFIX that does not also have the python stuff. Previously, it would just install half in /Users/leo/tmp2/bin/ and the python stuff elsewhere in /opt/homebrew/, I think.

Just sorting this out has probably already taken at least 30% of the Cadabra development time (with packaging for Windows the remaining 70%...) so it is unlikely that I wlll find the energy to give this another attempt soon.

If you just want to test, create a Python virtual environment and install Cadabra while you have that active.

Answered: cadabra2python seg fault

Tue, 08 Apr 2025 09:39:48 +0000

Now fixed in the devel branch.

Answered: curly-braces and lost indices

Tue, 08 Apr 2025 07:34:53 +0000

Yes, the general advice is to not sprinkle curly braces around if you don't need them. I agree though that this is unexpected and should at least throw an error. I have added this to the issue tracker so it doesn't get lost, https://github.com/kpeeters/cadabra2/issues/355.

Answered: simple substitute hits wrong target

Tue, 08 Apr 2025 07:28:47 +0000

You smash a bug and it appears somewhere else... This one was caused by fixing a related issue just after X-mas (I blame Santa). Hopefully it is now finally gone. Pushed a fix to the devel branch. Thanks for reporting this!

Answered: Evaluate does not respect NonCommuting

Fri, 14 Mar 2025 07:31:34 +0000

Evaluate calls simplify under the hood. You can turn this off by using

evaluate(_, simplify=False);

However, I think you probably want

product_rule(_);

instead (evaluate is for evaluating components of a tensor expression, but yours is a scalar, so it is already 'evaluated' for that purpose; see the docs for more detail and ask again if it's not clear).

Answered: How to manage two projects in their own folders which import .cnb's with the same names

Mon, 17 Feb 2025 14:23:38 +0000

The version of Cadabra in the devel branch on github contains code which avoids multiple identically-named notebooks mapping to the same converted .py file. You can build this from source or wait a little until this gets released as 2.5.12.

Answered: No GUI after compiling from the source (manjaro)

Thu, 02 Jan 2025 15:11:47 +0000

That sounds you either have some old version of cadabra-server floating around, or there is a mixture of different python versions on your system which are not picked up consistently. Can you email me the output of the cmake stage (with a clean build folder)?

The theme warning errors probably have something to do with customisations you (or some manjaro package) has made to the gtk theme.

Answered: evaluate() returns RuntimeError when using the explicit value of an index

Fri, 27 Dec 2024 23:06:02 +0000

This is now fixed on the devel branch which will become 2.5.12 eventually.

Answered: Canonicalise Kernel crash

Fri, 27 Dec 2024 11:57:57 +0000

The presence of those squares (e.g. (\partial_{t}{p})**2 trips canonicalise. In general, this algorithm cannot deal with powers; you have to write them out first with expand_power if you want this to work. I will add a test to canonicalise to prevent it from attempting to canonicalise these expressions.

Answered: derivatives are not evaluated in evaluation of components

Thu, 26 Dec 2024 19:57:32 +0000

After your substitute(ex2, cc), you should simply be able to do

evaluate(ex2, join(metricRdp2, global2));

This evaluates the components of the expression ex2 using the rules in metricRdp2 and global2.

Answered: compile failed using boost 1.87

Tue, 24 Dec 2024 11:18:24 +0000

2.5.10 is now available via homebrew; please try and let me know if there are any more issues.

Answered: c++lib fails to compile

Wed, 11 Dec 2024 08:45:09 +0000

Thanks! I have pushed changes for this to the devel branch, and have also added it to the CI pipeline so the c++lib build process is tested on every commit; hopefully that will prevent this from getting out of sync again.

There are not many people that I am aware of who use Cadabra's c++lib. Care to say a few words about what you are trying to do with it?

Answered: Kronecker delta wrong output in 2D

Tue, 10 Dec 2024 21:05:35 +0000

Ouch. The code responsible for this bug was about 10 years old, did not have the appropriate tests, and indeed produced nonsense when the range of the integer was symbolic. Thanks for reporting this.

I have pushed a fix to the devel branch on github; this will go into 2.5.10 to be released soon.

Answered: How to handle metric on submanifold?

Sat, 07 Dec 2024 15:21:13 +0000

I can't reproduce that; are you sure that with a fresh kernel start the above minimal example does the wrong thing? If so, have you tried with the current version (2.5.8 or the upcoming 2.5.10 which is available as 2.5.10-rc1)?

Answered: collect_factors / simplify with AntiCommuting objects

Thu, 21 Nov 2024 20:39:59 +0000

That simplify butchers the expression is expected, as no info about (anti-)commutativity currently gets passed through the sympy bridge. However, collect_factors does not use sympy and should know better. I have opened an issue on github to keep track of this, https://github.com/kpeeters/cadabra2/issues/324 .

Cadabra in Jupyter notebook - how to store?

Sun, 20 Oct 2024 12:59:00 +0000

Hello, I'm working with Cadabra on Jupyter notebooks, specifically via JupyterHub running on a linux system. I save my work and shuts down the computer until my next session, but then when I open the notbook again I see that the results are fonw, only the code is saved. I think only once I managed to somehow save a notebook with all of the results still showing when I re-open it. I don't know how I did it.

So my questio nis how to save a Cadabra-notebook (on a jupyterhub) with all the data still in the cache, and saved even after the computer is shut down?

Answered: sympy(?) errors parsing nested derivatives?

Tue, 01 Oct 2024 07:23:06 +0000

That problem comes from the fact that you wrote \sin\phi, instead of \sin(\phi). In Cadabra you have to group function arguments. (Of course I agree the error is useless and this should be caught earlier).

Answered: \partial_\theta causes infinite loop(?) in server

Mon, 30 Sep 2024 09:58:40 +0000

I guess you have your answer there: add braces. In general, Cadabra is less forgiving about not wrapping arguments in braces, so just do it always.

Cadabra should not hang of course, will investigate. I have opened an issue at https://github.com/kpeeters/cadabra2/issues/312.

Answered: Custom display symbols for variables?

Mon, 30 Sep 2024 07:19:15 +0000

Yes, use LaTeXForm, as in

vr::LaTeXForm("\dot{r}").
vtheta::LaTeXForm("\dot{\theta}").

ex:= vr + vtheta;

gives $\dot{r}+\dot{\theta}$.

Answered: Gtk error messages on startup.

Mon, 30 Sep 2024 07:15:18 +0000

I may need to see if those gtk modules should be included in the AppImage too (I have not done a lot of testing with the AppImage on systems that are truly bare, without an existing gtk installation).

The connection to server failed message happens when you restart the kernel or load a new notebook (which restarts the kernel too). This kills the existing cadabra-server first, and the client then reports that. I could probably remove that error as it sounds more alarming than it is.

Answered: Simple derivative evaluation not working for me.

Mon, 30 Sep 2024 07:12:55 +0000

The evaluate does not substitute scalar functions, only component values (which is arguably misleading, and will possibly change in the future). But you can easily work around this by first substituting and then evaluating:

substitute(exfr, exf);
evaluate(exfr);

In fact the 2nd line then only really calls simplify, so you may as well use that.

Answered: Another "No module named 'sympy' " question.

Sun, 29 Sep 2024 20:58:15 +0000

I have just released 2.5.6, which fixes this issue, and also includes plotting support in the AppImage version. You can find the updated AppImage in the github release assets for this release.

Answered: Cycle freeze

Thu, 26 Sep 2024 17:01:38 +0000

I cannot reproduce this. Which "today's update" are you talking about? I did not make any new release. What is the version and build number? (start the command-line cadabra2 and look at the first line of output).

Answered: Expanding sum into time and purely space components.

Mon, 12 Aug 2024 09:46:46 +0000

Hi Heckfy.

This "anwser" is more a comment, but it might transform onto an answer after some discussion.

First contact

I understand what you are trying to achieve, but not your question specifically. Let me comment further.

There are typographic in your code blocks, and missing components in your substitution rules (non-diagonal inverse components of the background metric).

I think that the right substitution rule should be

back := gb_{t t}=-1, 
  gb_{r r}=a**2/(1-k*r**2), 
  gb_{\theta \theta}=a**2*r**2, 
  gb_{\varphi \varphi}=a**2*r**2*\sin{\theta}**2, 
  gb_{t r}=0, 
  gb_{t \theta}=0, 
  gb_{t \varphi}=0, 
  gb_{r \theta}=0, 
  gb_{r \varphi}=0, 
  gb_{\theta \varphi}=0,
  gb_{r t}=0, 
  gb_{\theta t}=0,
  gb_{\varphi t}=0,
  gb_{\theta r}=0,
  gb_{\varphi r}=0, 
  gb_{\varphi \theta}=0
  gb^{t t}=-1, 
  gb^{r r}=(1-k*r**2)/(a**2), 
  gb^{\theta \theta}=1/(r**2 * a**2), 
  gb^{\varphi \varphi}=1/(a**2 * r**2 *\sin{\theta}**2), 
  gb^{t r}=0, 
  gb^{t \theta}=0, 
  gb^{t \varphi}=0, 
  gb^{r \theta}=0, 
  gb^{r \varphi}=0, 
  gb^{\theta \varphi}=0,
  gb^{r t}=0, 
  gb^{\theta t}=0,
  gb^{\varphi t}=0,
  gb^{\theta r}=0,
  gb^{\varphi r}=0, 
  gb^{\varphi \theta}=0;

In your question you talk about the connection and Ricci tensor, but there is nothing in the presented code referring to those objects. Could you please update the post to include a more complete notebook?
The second code block in your post does not compute anything! Remind that cadabra is not a calculation software, but a manipulation software. That means that you have to tell it explicitly what to do. So far (at least in what you showed), it is doing nothing.

Well, in short, could you improve the quality of your post: improve the statement of your question, and provide a more complete version of your code.

Cheers, Dox.

Answered: question about eliminate_metric

Mon, 05 Aug 2024 14:26:20 +0000

Hi Eureka and Kasper.

Thank you K for the suggestions in solving the issue. I used your point to evaluate the code by Eureka.

I noted that the accent desappear when the ex got evaluated, unless the accent surround just the symbol g (wihtout the indices).

Here my code:

{\mu,\nu,\rho}::Indices(vector, position=fixed).
g^{\mu\nu}::InverseMetric.
g_\mu^\nu::KroneckerDelta.
g^\mu_\nu::KroneckerDelta.
\delta{#}::Accent.

ex:=\delta{g}_{\mu\nu} g^{\nu\rho};
eliminate_metric(_);

Best wishes, Dox

Answered: A bug of Accent

Thu, 01 Aug 2024 08:51:17 +0000

Hi Eureka.

Firstly, let me point that a semi-colon is missing at the en d of the third line. So, the correct code should be

{a,b}::Indices(vector).
\delta{#}::Accent.
ex := \delta{A^{a b}+B^{a b}};

Then, let me clarify that (to my understanding) accents are a way to differentiate symbols, i.e. it has no further properties. Why is this important? Perhaps your expression should be written as

ex := \delta{A}^{a b} + \delta{B}^{a b};

If you are expect your accent to have linearity property, give assign the property of Derivative, i.e.

\delta{#}::Derivative.

If you use the Derivative property instead of the Accent, you could apply the distribute algorithm in your original expression

{a,b}::Indices(vector).
\delta{#}::Derivative.
ex:=\delta{A^{a b}+B^{a b}};
distribute(ex);

Cheers, Dox.

Answered: Factor out of numerical coefficient

Thu, 18 Jul 2024 13:30:43 +0000

The main reason why you cannot factor out only a numerical factor is that Cadabra stores rational pre-factors in a special location (the 'multiplier') inside each node in the expression tree.

Cadabra will always rewrite expressions so that a \sum node will have a unit multiplier. So all multipliers are associated to the summands. This is a canonicalisation choice which many algorithms in Cadabra rely on, so it's not easy to change that. Similarly, all child nodes of a \prod node will have unit multiplier.

As soon as you have a product in your expression node, like in 6 A (B + C), the multiplier is associated to the \prod node. So you still have

\prod{A}{ \sum{B}{C} }

with a unit multiplier for the \sum, but now the \prod node has multiplier 6.

Answered: I'm trying to reproduce equation 4.6 from 4.5 in Carrolls Spacettime ;and Geometry

Fri, 28 Jun 2024 08:24:56 +0000

Hi ChuckW.

I ran your code, compared the output with the result reported in the book, and it seems that the result is correct.

The last piece of the puzzle is the use of the chain rule. I complemented your code with the following:

chain_rule := \partial_{\lambda}{x^{\nu}} \partial_{\nu}{A??} -> \partial_{\lambda}{A??};

substitute(expr, chain_rule);

Hope this suggestions works to complete your task.

Cheers, Dox.

how to deal with lovelock gravity?

Sun, 23 Jun 2024 03:15:45 +0000

For Lovelock gravity,

$$ L=R_{a_{2}b_{2}}^{c_{2}d_{2}}\cdots R_{a_{m}b_{m}}^{c_{m}d_{m}}\delta_{c_{2}d_{2}\cdots c_{m}d_{m}}^{a_{2}b_{2}\cdots a_{m}b_{m}} $$

where

$$ \delta_{{jc_{1}d_{1}...c_{m}d_{m}}}^{{ia_{1}b_{1}...a_{m}b_{m}}}=\det\left[\begin{array}{c|ccc}\delta_{j}^{i} & \delta_{{c_{1}}}^{i} & \cdots & \delta_{{d_{m}}}^{i}\\hline \\delta_{j}^{{a{1}}}\\vdots & & \delta_{{c_{1}d_{1}...c_{m}d_{m}}}^{{a_{1}b_{1}...a_{m}b_{m}}}\\delta_{j}^{{b\{m}}}\end{array}\right] $$

when $m=2$, we get Einstein gravity; when $m=3$, we get Gauss-Bonnet gravity. But how to deal with it more generally? More concretely, I don't know how to do it use for loop.

LaTeXForm fails in version 2.5.2

Mon, 17 Jun 2024 22:50:17 +0000

Greetings,

This simple statement

Dx{#}::LaTeXForm{"Dx"}.

raises a syntax error

SyntaxError: '{' was never closed

I'm running Cadabra version 2.5.2. build 3043.2a1d62739a.

Also, this version of Cadabra compiles fine under MacOS Sonoma 14.5.

Cheers, Leo

substitute gives incorrect value on scalar expression

Mon, 17 Jun 2024 22:25:46 +0000

Greetings, I'm trying to use the substitute algorithm to set the value of a scalar expression but I'm getting some very odd results. In the following examples I set a variable s to be 1 and the expected result should be zero. But in the first case I get 2. This problem only seems to arise for the single choice of s=1. Cheers, Leo

# ----------------------------------------------------
# fails, should return 0

foo := s - s**3;

substitute (foo, $s = 1$);

# ----------------------------------------------------
# success

foo := s - s*s*s;

substitute (foo, $s = 1$);

# ----------------------------------------------------
# success

foo := s**3 - s;

substitute (foo, $s = 1$);

# ----------------------------------------------------
# success

foo := s - s**3;

substitute (foo, $s = 2$);

# ----------------------------------------------------
# success

foo := s - s**3;

substitute (foo, $s = 1/2$);

Cadabra 2.5.x released

Sat, 15 Jun 2024 19:27:27 +0000

I have today taken Cadabra 2.5.x out of experimental mode and made it the default.

This version contains built-in code to handle LaTeX typesetting, making the notebook interface much faster and enabling things like dynamical cell updates (e.g. updating counters) as well as full UTF-8 support for non-English languages (and emojis...). In the process, I have also fixed a number of issues with the notebook, e.g. interrupting calculations.

As a bonus, it is now possible to distribute Cadabra as an AppImage on Linux, which installs on any reasonably modern distribution. Hopefully this will help to reach a larger audience. I have also updated the homebrew package for macOS. Standard .deb packages for Ubuntu 22.04 are available as well.

Any feedback is welcome; please try this one and let me know!

location of bracket blocks distribute

Tue, 11 Jun 2024 05:46:21 +0000

Hi Folks,

Here are two instances of the distribute algorithm. Only the first gives the expected result. The only difference in the code is where the opening bracket is placed.

# ----------------------------------------------
# this works as expected

foo := A (
       B + C);

distribute (foo);

# ----------------------------------------------
# this fails to apply the "distribute" algorithm

bah := A
       (B + C);

distribute (bah);

Cheers, Leo