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When people are thinking, most of them have inner monologues. However, I'm facing problems because I don't know how to properly process Math expressions in my inner monologue. For example,

$$\mathrm{Hom}_{\mathcal{O}_X}(f^*\mathcal{G},\mathcal{F})\cong \mathrm{Hom}_{\mathcal{O}_Y}(\mathcal{G},f_*\mathcal{F})$$

or

$$\mathrm{id} \times \psi^{\otimes n}:\mathcal{F}\otimes \mathcal{L}^n \vert_U \cong \mathcal{F}\vert_U$$

Edit: There might be no canonical inner monologue, but I would like to gain some experience from others, since I am often anxious because I think mine sucks.

Moreover, if you speak multiple languages, in what language do you have numbers in your inner monologue?

I'm also curious about how do people without inner monologues think about Mathematics.

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    $\begingroup$ What makes you think there is a ‘proper’ monologue associated to a given expression? It’s definitely the case that different mathematicians have different experiences and perspectives on the same ideas, different ways to encode their intuitions and explore the subject; good! Without that, mathematics would hardly progress $\endgroup$
    – FShrike
    Nov 20, 2022 at 13:07
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    $\begingroup$ @FShrike There's not a canonical inner monologue, but I find mine not very satisfying. I would like to see how other people encode their thoughts. For example, how do you encode $\in$ and $\subset$ in your monologue? $\endgroup$ Nov 20, 2022 at 13:18
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    $\begingroup$ What makes you dissatisfied with your inner encoding of $\in$ and $\subset$? You should come up with something more juicy. A student of Otto Forster told me once that Professor Forster used to say when looking at certain papers: "I can already see from the typesetting that this must be wrong." $\endgroup$
    – Kurt G.
    Nov 20, 2022 at 13:29
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    $\begingroup$ @KurtG. In my mother language Chinese, $A \in B, A \subset B$ can be pronounced shortly as "A shuyu B, A hanyu B". But when I'm reading English material, I'd like to keep the inner encoding of symbols uniform (no Chinese words appearing in English context), short, and of infix notation. I'm looking for such a representation. $\endgroup$ Nov 20, 2022 at 13:44
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    $\begingroup$ Some don't have monologues, some have visualizations especially when the problem is geometrical in nature. The fact of the matter is, "monologue" is the product of ones own intuition, not the other way around. What you should be asking is how to build that intuition $\endgroup$ Nov 20, 2022 at 14:07

4 Answers 4

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Expressions I am very familiar with usually don't translate into internal monologue for me, it feels more like I am sensing the underlying mathematical statement directly. For example, in $\int_a^b f(x) dx$ I don't feel compelled to put any linear order on the information - I perceive bounds, function and integration variable/measure in parallel.

If there is internal monologue, it is going to be in English (which is not my mother tongue), and may involve LaTeX-commands to "read" symbols.

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    $\begingroup$ I share this perception. Seeing the integral, my brain thinks "area under the curve". $\endgroup$ Nov 20, 2022 at 22:02
  • $\begingroup$ I also share your perception. Also, if I read mathematics outside the community where it is spoken I adopt an "intermediate language" (maybe involving Latex) as long as I not received the phonetics from a member of this community. When done, my brain (in general) automatically adjusts. $\endgroup$ May 16 at 16:26
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For what it's worth, here is a redacted version of my inner monologue upon being exposed to your first example

$$\mathrm{Hom}_{\mathcal{O}_X}(f^*\mathcal{G},\mathcal{F})\cong \mathrm{Hom}_{\mathcal{O}_Y}(\mathcal{G},f_*\mathcal{F})$$

"So, it's two hom-sets that are basically equal, which ones? Oh wait, first note they are different kinds of homs, one is sort of over $X$ and the other over $Y$; ah I see, there is probably a map -- yup, that's the $f$ inside there -- and probably something gets pushed forward and/or pulled back with that map. Ah, I see, it's those sheaf things, never liked them, but basically I guess it's like modules, so let's think of $\mathcal F$ and $\mathcal G$ as modules, and there's a map $f$ between, uh, either those modules or their rings or something. Ah, I guess $\mathcal{G}$ is a $Y$-module and $\mathcal{F}$ is an $X$-module, and the $f^*$ turns $Y$-into $X$-modules and the $f_*$ the other way. And somehow that is probably something extremely natural when one writes it down, that some homomorphisms of $X$-modules can naturally be translated into homs of $Y$-modules, and back. Or at least one direction in these things is always trivial, maybe the other needs work, but it's probably not worth writing it down. So some hom-set gets identified with another hom-set, over a different ring, by abstract nonsense, probably helpful changing from one ring to another and back."

And as for

$$\mathrm{id} \times \psi^{\otimes n}:\mathcal{F}\otimes \mathcal{L}^n \vert_U \cong \mathcal{F}\vert_U$$

"What the ... ugh, that's shit notation out of context, the whole first thing has to be read together, one should put parentheses around it, like this: $(\mathrm{id} \times \psi^{\otimes n})$. And that is a map, yes? From, hmm, that tensor product, whatever that is, but obviously $id$ acts on the first factor, and on the second, that kind of $n$-fold product of $\psi$, which makes sense as $\mathcal{L}^n$ must be some $n$-th power, although now I don't understand that tensor power in the map. Whatever. Oh and it says it makes something isomorphic to just the $\mathcal{F}$, on $U$, whatever that all is. So it seems like that tensor power of that $\psi$ sort of kind of kills the second factor? Alright. Would need to look closer into this."

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    $\begingroup$ “Ah, I see, it's those sheaf things, never liked them, but basically I guess it's like modules” ­– haha, I don't know how many times I've thought almost exactly that! $\endgroup$ Nov 22, 2022 at 14:07
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    $\begingroup$ I am not sure that “this is useful” in the sense of helping the questioner to experience notation more effectively, but ▲ anyway, as it is a frank statement of what happens in your head, which is what they asked for, and as such could help a discussion along. $\endgroup$
    – PJTraill
    Nov 24, 2022 at 15:12
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I agree with Arno: inner monologue isn't actually the right approach to maths. (For that matter, IMO it's also not that great for many other things... I'd estimate that only about 25% of my thoughts go with an inner monologue.) Much better is to associate formulas with pictures.

When I read something about Hom-sets, sketches of categories and arrows turn up in my brain. When I read about isomorphism, a little animation starts playing that uniquely connects pairs of elements of the two sets. When I read about cartesian products, something folds up like a book to reveal another dimension. With tensor products... actually hard to describe, but again it taps into the learned geometric intuition, not into any communication in words.
You get the general idea. Associating all these things to words isn't really useful, except insofar as it's good training in order to be able to better explain things on a blackboard. What you remember by name are theorems (and definitions / axioms), not formula symbols / syntactic expressions.

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I interpret this as a question posed by a non-native speaker about how people normally pronounce common mathematical symbols in English, rather than a nebulous question about inner monologues. Here are some acceptable pronunciations of common mathematical symbols.

For each $a \in X$ [$a$ in $X$; element $a$ of $X$].

It follows that $a \in X$ [$a$ lies in $X$; $a$ is in $X$; $a$ belongs to $X$].

For each $a \leq b$ [$a$ below $b$].

It follows that $a \leq b$ [$a$ is below $b$; $a$ lies below $b$].

For each $a \geq b$ [$a$ above $b$].

It follows that $a \geq b$ [$a$ is above $b$; $a$ lies above $b$].

For each $X \subseteq Y$ [subset $X$ of $Y$; $X$ subset of $Y$; $X$ contained in $Y$].

It follows that $X \subseteq Y$ [$X$ is a subset of $Y$; $X$ is contained in $Y$].

$A \otimes B$ [$A$ tensor $B$]. $A \times B$ [$A$ cross $B$].

$f^{-1}$ [$f$ inverse]. $f^{*}$ [$f$ star].

$\mathrm{Hom}(A, B)$ [Hom $A$ $B$; Hom of $A$ $B$; Homset of $A$ and $B$].

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    $\begingroup$ How to differentiate between $f^*$ and $f_*$? Moreover, how to pronounce $f^\#$? $\endgroup$ Nov 21, 2022 at 5:39
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    $\begingroup$ I don't like "$a$ below $b$" (that would be $a<b$). Better "$a$ up to $b$" or "$a$ at most $b$", or the full "$a$ lesser or equal $b$". $\endgroup$ Nov 21, 2022 at 8:21
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    $\begingroup$ @AprilGrimoire you don't differentiate between $f^*$ and $f_*$ unless it is really ambiguous. (In which case I would question whether that was a good choice of notation in the first place!) In your example, it's disabiguated by type of what you apply it to. As for $f^\#$, you could pronounce that "$f$ hash" (or "pound"), but preferrably use whatever higher-level concept the # mark represents in this particular application instead. $\endgroup$ Nov 21, 2022 at 8:25
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    $\begingroup$ In contexts where $f^\sharp$ and $f^\flat$ appear, you might also read them as “$f$ sharp” and “$f$ flat”, like in music. (I’m always tempted to use the German versions which are “-is” and “-es”, even when they’re applied to things that are not also names of musical notes: “Kes” or “Q-p adjungiert alle p-ten Wurzeln aus p-es” …) $\endgroup$ Nov 21, 2022 at 8:39
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    $\begingroup$ @leftaroundabout I agree, but there are some historical conventions, for example the direct image and inverse image of schemes. $\endgroup$ Nov 21, 2022 at 8:50

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