Sorry for the terse title; the 150-character limit really gets in the way when you have to include "\varepsilon" in the title several times.
Write $k[\varepsilon]$ for the dual numbers, i.e. $k[x]/(x^2)$.
If we glue two copies of $\operatorname{Spec} k[\varepsilon]$ along their closed points, w...
It seems to me that this is a phenomenon that even happens in differential geometry. Really, unless you make an explicit identification of $T_0\mathbb{R}^2$ with $\mathbb{R}^2$ there is no real labeling of the axes--this 'directionality' of tangent vectors in terms of the original space is somewhat choice dependent no? The space of tangent vectors doesn't contain knowledge about actual directions on the original space, but what it does contain is 'independence of directions'. I can say more if this resonates at all with you.
"Really, unless you make an explicit identification of $T_{(0,0)} \mathbb{R}^2$ with $\mathbb{R}^2$ there is no real labeling of the axes" Well, curves through $(0,0)$ in $\mathbb{R}^2$ determine tangent vectors in $T_{(0,0)} \mathbb{R}^2$. That doesn't require making any extra identifications.
Like what I'm saying to you is that $T_0\mathbb{R}^2$ is $\mathbb{R}^2$ just like $\mathcal{O}_{X,0}/\mathfrak{m}_0^2\cong k[\varepsilon_1,\varepsilon_2]$ (where $X$ is the intersecting lines). Once you choose an identification of $\mathbb{R}^2$ with $T_0\mathbb{R}^2$ you can make sense of specific directions, but not before that.
How are you defining $T_{(0,0)} \mathbb{R}^2$ in such a way that it doesn't come preequipped from the outset with a map taking curves through $(0,0)$ in $\mathbb{R}^2$ to vectors in $T_{(0,0)} \mathbb{R}^2$?
Let $V$ be a two-dimensional vector space. Then, $V$ does not inherently have directions. But, an identification $V\cong T_0\mathbb{R}^2$ allows one to give a mapping from curves through $0$ in $\mathbb{R}^2$ to $V$ giving $V$ a sense of direction. I'm saying that $\mathrm{Spec}(k[\varepsilon_1,\varepsilon_2])$ is like $V$ and you can give it 'explicit directions' by choosing an identification of $k[\varepsilon_1,\varepsilon_2]$ with $\mathcal{O}_{X,0}/\mathfrak{m}_0^2$. Does that clarify at all what I'm trying to communicate? If not, I'll give some thought of how to say it more cogently.
@AlexYoucis I agree with that but I think the situation at hand is different. $\operatorname{Spec} k[x,y]/(xy)$ inherently has two privileged directions; $\operatorname{Spec} k[\varepsilon_1, \varepsilon_2]$ does not. But both are formed by gluing "lines" at a point; it's just that one is an actual line and the other is an "infinitesimal line."
So if you take two A^1s, glue them at a point, then forget the inclusion maps of the original A^1s, then the resulting space still has privileged directions
For example if you take the tangent cone at the singular point it is not the entire tangent space
However if you take two Spec k[e]s and glue them at the closed point, forgetting the inclusion maps from the Spec k[e]s
Then you have no way to distinguish any two vectors in the tangent space to the closed point
This is weird to me, which I guess means I don't really grok what a tangent space is, and least outside the smooth case.
@DanielMcLaury OK, I think now we're changing the question in some sense. Are you asking about the space $\mathrm{Spec}(k[\varepsilon_1,\varepsilon_2])$ and its relation to the space $\mathrm{Spec}(k[x,y]/(xy))$ or theirtangent spaces?
I feel like you're using directions in a sort of amorphous way.
Discussion between Daniel McLaury and…
Discussion between Daniel McLaury and…