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I am currently reading an article written by John Mather called "Stability of $C^{\infty}$ mappings III. Finitely determined map-germs". I am note very familiar with algebra (basic one semester curse) so I hope someone can help me to identify the object in question (see picture below, Corollary 3).

For better understanding: $C_S=C(N)_S$ denotes the set of $C^{\infty}$ map-germs $(N,S) \to (\mathbb{R}, \mathbb{R})$ and $m_S=m(N)_S$ denotes the ideal in $C_N$ consisting of $C^{\infty}$ map-germs $(N,S) \to (\mathbb{R},0)$.

First question: does $m_S^k = m_Sm_S \cdot \cdot \cdot m_S $ (k-times) mean the product of group subsets?

Second question: The object in corollary 3: $C_x / m_x^k$ does it mean the quotient group or the quotient ring of the two objects?

EDIT: The more I think about it, I am sure that it is the quotient ring. But how can I imagine this set? How does the set look like? enter image description here

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The subset $m_S^k$ consists of the finite sums of elements of the form $f_1f_2\cdots f_k$, where $f_1,\ldots,f_k\in m_S$. In other words, it is the smallest ideal containing all the $f_1f_2\cdots f_k$. Since I do not understand what product of group subsets means, can't say more here.

Concerning the last question, indeed, $C_x/m_x^k$ is the quotient of the ring $C_x$ by the ideal $m^k_x$. If you feel more comfortable with groups: if we forget the multiplication, then this quotient ring is actually the quotient group of $C_x$ by the subgroup $m_x^k$ (where, however, $m_x^k$ is still defined as above!)—that is, the elements are equivalence classes of elements of $C_x$, i.e., germs at $x$, where two germs are equivalent if their series approximations equal in degree $<k$.

Let's take $n=1$ for example. Then $m_x$ is just the set of germs of functions vanishing at $x$. In other words, it is the kernel of the ring map $C_x\to\mathbb{R}$, $f\mapsto f(x)$. Since this map is clearly surjective, it induces an isomorphism (of rings) $C_x/m_x\to\mathbb{R}$. This is just the first case of Corollary 3. In general, the ring $C_x$ is rather complicated; this is why Corollary $3$ is important. It tells us that up to finite order, it's just power series up to finite order. Let's explore the right hand side of the isomorphism, $\mathbb{R}[[x_1,\ldots,x_n]]/m^k$, where $m = (x_1,\ldots,x_n)$ and so $m^k=(x_1^k,x_1^{k-1}x_2,\ldots \text{ all monomials of degree }k\ldots, x_n^k)$. With a power series $f = \sum f_Ix^I$ (multi-index notation), we may associate a polynomial $\tau^{<k}(f) := \sum_{\#I<k}f_Ix^I$ where we simply forget all terms of degree greater or equal than $k$. This gives a ring map $\tau^{<k}\colon \mathbb{R}[[x_1,\ldots x_n]]\to \mathbb{R}[x_1,\dots x_n]$ whose kernel is precisely $m^k$; thus, it induces an isomorphism between $\mathbb{R}[[x_1,\ldots,x_n]]/m^k$ and the ring of polynomials in $n$ variables of total degree less than $k$.

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  • $\begingroup$ Very good, thanks you! English is not my mother tongue so I checked wikipedia what the english word is and as you described it, it turns out that it is exactly what I had in mind. Here is the link for you to check: en.wikipedia.org/wiki/Product_of_group_subsets For better understanding: In general would it be better to work with the right hand side of corollary 3 rather than with the left hand side? For example if I am interested in the tangent space of $C_x/m_x$. $\endgroup$ – JDoe Sep 28 '16 at 7:38
  • $\begingroup$ Well, it's a ring, so maybe you don't want to talk about the tangent space of $C_x/m_x$, but the tangent space of $N$ at $x$. So, a tangent vector should be a first order direction at $x$, right? Unfortunately, $C_x$ does not know directions. However, a 'first order direction' is uniquely determined by its value on 'first order functions'. This can be made precise in the sense that the tangent space at $x$ is the $\mathbb{R}$-vector space dual to $m_x/m_x^2$. This usually goes under the name Zariski tangent space if you want to look it up. $\endgroup$ – Ben Sep 28 '16 at 13:30
  • $\begingroup$ I will check this definitely. In case I still have questions about this topic, can I post them here although the question might not fit the starting question? $\endgroup$ – JDoe Sep 28 '16 at 14:28
  • $\begingroup$ Strictly speaking, we're not supposed to do that in comments. I'd like to help you, though, so you could either ask here and we'll see if it's more appropriate for a new question, or you could just ask a new question and let me know that you did via a comment here. A new question has the advantage that it may also attract others. $\endgroup$ – Ben Sep 28 '16 at 14:46
  • $\begingroup$ Alright I will start a new question and let you know in the comments here. Thanks! $\endgroup$ – JDoe Sep 28 '16 at 14:54

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