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In Anders Kock's book Synthetic Differential Geometry, the following objects are introduced (I used brackets but the notation stands for the corresponding object in your favorite $$D_k(n)=\left\{ (x_1,\dots ,x_n)\in R^n\mid \prod_{i=1}^{k+1}x_{\ell _i}=0 \; \forall \ell_i \right\}$$ where $R$ is the geometric line. So $D_k(n)$ is comprised of all $n$-tuples such that the product of every $(k+1)$-tuple is zero. These objects are shown to have some nice formal properties, but I don't understand what their "elements" mean geometrically. Elements of $R$ satisfying $x^2=0$ I can understand as tiny, so the definition of the tangent bundle as an exponential is very clear.

How to make sense of the $D_k(n)$'s?

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    $\begingroup$ What do you mean by $\prod_{k=1}^{k+1}$? You seem to be using $k$ for two different things. Did you maybe mean $\prod_{i=1}^{k+1}$? Also, what do you mean by $\forall \ell_i$? Do you mean for all $k+1$-tuples $(\ell_1,\dots,\ell_{k+1})$, where each $\ell_i \in \{1,\dots,n\}$? $\endgroup$ – Nick Feb 22 '16 at 22:03
  • $\begingroup$ @Nick yes on all accounts. I edited the question. $\endgroup$ – Arrow Feb 23 '16 at 8:40
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You might call $D_k (n)$ the $n$-dimensional infinitesimal disc of order $k$. In particular, note that $D_1 (1)$ is (isomorphic to) $D = \{ x \in R : x^2 = 0 \}$, and in general, $D_k (1)$ is (isomorphic to) $D_k = \{ x \in R : x^{k+1} = 0 \}$. Note also that $D_1 (1) \subseteq D_2 (1) \subseteq D_3 (1) \subseteq \cdots$: so "higher-order" infinitesimals are, in a sense, larger rather than smaller.

It's a bit harder to explain why $D_k (n)$ is defined the way it is for $n \ge 1$. Obviously, $D_k (n) \subseteq (D_k)^n$, but why do we include the condition on cross-terms? I think the answer is that these definitions are designed with the theory of power series in mind. For instance, $D_1 \subseteq R$ is not closed under sums, but for $(x_1, \ldots, x_n) \in D_k (n)$ and $(\lambda_1, \ldots, \lambda_n) \in R^n$, we have $\lambda_1 x_1 + \cdots + \lambda_n x_n \in D_k$, i.e. $(\lambda_1 x_1 + \cdots + \lambda_n x_n)^{k+1} = 0$. In fact:

Let $\vec{x} \in R^n$. The following are equivalent:

  • $\vec{x} \in D_k (n)$.
  • For every $m$-linear form $\phi : R^n \to R$, if $m > k$, then $\phi (\vec{x}) = 0$.

Thus, it makes sense to evaluate power series at elements of $D_k (n)$, and moreover, the value is equal to the truncation of the power series to terms of total degree $\le k$.

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  • $\begingroup$ I also thought that geometrically the condition $x_i\cdot x_j=0$ for all elements $(x_1,\dots ,x_n)\in D(n)$ is analogous to how $dx \cdot dy =0$ because both, while different infinitesimals, are so small that their product is zero. Does this make sense? $\endgroup$ – user153312 Feb 27 '16 at 18:50
  • $\begingroup$ Sure. That part is obvious. $\endgroup$ – Zhen Lin Feb 27 '16 at 18:52

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