What inner monologue appears when you read Mathematical expressions?

Question

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.

Answer 1

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.

Answer 2

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 $B$].

Answer 3

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.

Answer 4

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."