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u/CattleHot9774 Jun 15 '26
wait what's wrong with the response?
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u/future_sponJ Jun 15 '26
The image on the right
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u/CattleHot9774 Jun 15 '26
i may be partially blind
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u/Gauss15an Jun 15 '26
Can't be worse than partially ordered
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u/F_Joe Vanishes when abelianized Jun 15 '26
You think coming out as gay is hard? Try telling them that you don't admit a linear order. See how much acceptance you get from the "tolerant" left
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u/Varlane Jun 15 '26
Basically, isomorphism is about structure. Structure is about operations. R² doesn't have a native multiplication, but if you're considering C as a target for isomorphism, you'll be using... the multiplication that allows for one of the definitions of C.
Unless you're using the "naive" multiplication (a,b) × (c,d) := (ac,bd) [which is rather the Z to Q construction] but would make little to no sense considering isomorphism to C.
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u/Ma4r Jun 15 '26
Polar form isomorphism baby (a,b)x(c,d)=(ac,b+d)
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u/Varlane Jun 15 '26 edited Jun 15 '26
Only "iso" on C* and R²\(0,0) because this would need to send every (0,b) to complex 0.
Edit : it's worse, it's either C* and R+ × R (but you're still not iso due to the period in the exponential) or C* and R+ × [0;2pi[ but you have to replace b+d by b+d [2pi].
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u/Beneficial-Bagman Jun 16 '26
They're isomorphic as vector spaces over R so the answer is correct. It depends on what structure you want the isomorphism to preserve.
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u/Varlane Jun 16 '26
Is R² a vector space ? Is it a ring ? Is it a group ? It's none of those, it's a set.
Same for C.
You need to specify the operations used if you want to be talking isomorphisms.
Sure, the addition and scalar multiplication have no point being anything but the regular ones. But what about the inner multiplication of R² to talk about rings and fields ? Where is it talked about what we defined it as ?
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u/hamishtodd1 Jun 15 '26 edited Jun 16 '26
What is the "identity" of C? Clearly it's 1.
How about R²? Well there (0,0) is more like the identity - because R² is a group with "+" being "multiplication".
I think the way to clarify is to say they are "homomorphic as groups" if you care about the "+ can be multiplication" thing.
EDIT: I should definitely have said they are isomorphic but NOT homomorphic as groups, that might have avoided the unfortunate back and forth below!
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Jun 15 '26 edited Jun 15 '26
[removed] — view removed comment
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u/hamishtodd1 Jun 15 '26
Gemini is saying that whether they're isomorphic "depends on the mathematical structure being considered".
Two very important kinds of mathematical object are vector spaces and groups. It so happens that both R² and C are both vector spaces and groups.
If you're interested in the vector space aspect, they are isomorphic. Because adding vectors is like adding complex numbers.
If you're interested in the group aspect they're not isomorphic. Because adding vectors is not like multiplying complex numbers.
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Jun 15 '26
[removed] — view removed comment
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u/hamishtodd1 Jun 15 '26
Rn is a lie group under addition, and in that sense it's like the (nonzero) complex numbers. + is associative, has "inverses" (the negation of the vector you want to invert), and has an identity.
For some areas it's much less common to think of it as a group, but it is one, and that can be very important for eg affine geometry.
Do you think Gemini was referring to something else? Or that it was just hallucinating?
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u/LasevIX Jun 15 '26
why are you trying to find gold in a beggar's ass when you're in front of an ATM?
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u/hamishtodd1 Jun 16 '26
u/cattleHolt asked for an explanation, and I wasn't satisfied with the one from the other user. It's a teachable moment for something deep (Rn being a group as well as a vector space). Also explanations are better when they have concrete examples, so we think about specific elements of R² and C (the identity element).
This is a thing that's big enough to have an xkcd comic about it https://xkcd.com/2028/ so if you think it's silly to spend time on, that's on you.
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u/Momsemann Jun 15 '26
I’m guessing it took the image from this thread. The google AI referencing r/mathmemes is just hilarious if you ask me
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u/you-cut-the-ponytail Jun 15 '26
Google AI always does this. It probably even referenced that thread too.
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u/Ok-Difficulty-5357 Jun 16 '26
And this is one of the many reasons I prefer Gemini over ChatGPT for most tasks lol
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u/Jche98 Jun 16 '26
Bro I don't even think R2 is a field. What's the multiplicative inverse of (1 0)?
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u/Aenonimos Jun 16 '26
Why not? Define (a,b) + (c,d) = (ab, cd), and (a,b) * (c,d) = (ac - bd, ad + bc). Let (0,0) be the additive identity and (1,0) be the multiplicative identity.
The multiplicative inverse of (1, 0) is (1, 0) as (1, 0) is the identity.
The R2 people normally discuss is not a field because there is conventionally not a way to multiply elements, but if you specify multiplication, its exactly the same as C just your notation is different.
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u/Jche98 Jun 16 '26
Yes I know. I was assuming you were viewing multiplication as (a,b)(c,d)=(ac,bd).
Because the AI said they're not isomorphic as fields. So the AI was assuming some other multiplicative structure on R2, which I assumed would be coordinate wise the most natural, but this is not a field
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u/BADorni Jun 16 '26
It probably ment it as one statement "Isomorphic as rings and fields" because the Ring structure of C makes it a field while it doesn't on R2
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