Quantum linear code/ Dual Code (CSS) proof

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Discussion Overview

The discussion revolves around the proof related to quantum linear codes and dual codes, specifically within the context of CSS (Calderbank-Shor-Steane) codes. Participants are exploring mathematical expressions and their implications in the theory of quantum error correction.

Discussion Character

  • Technical explanation

Main Points Raised

  • One participant requests clarification on the symbols used in the discussion, suggesting that more information or a reference to the relevant paper is necessary.
  • Another participant presents a mathematical argument regarding the case when a vector is not in the dual space, providing a series of equalities that rely on properties of subspaces.
  • A similar mathematical argument is reiterated by another participant, emphasizing the same points about the conditions under which the equalities hold.

Areas of Agreement / Disagreement

There is no clear consensus on the interpretation of symbols or the completeness of the proof, as one participant seeks clarification while others present mathematical reasoning without addressing the initial request for context.

Contextual Notes

The discussion may be limited by the lack of definitions for symbols and terms used, which could affect understanding and interpretation of the mathematical expressions presented.

steve1763
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TL;DR
What would the proof be for the following identity? I cannot find the proof anywhere
Screenshot 2021-09-05 at 21.52.33.png
 
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How should we know, what your symbols mean? You have to give a bit more information or a link to the paper you are referring too.
 
If ##x\in C^\perp## then it is clear. If not, then there is a ##c_0\in C## such that ##x\cdot c_0 =1##. The you have

##
-1\sum_{c\in C}(-1)^{x\cdot c}=(-1)^{x\cdot c_0}\sum_{c\in C}(-1)^{x\cdot c}=\sum_{c\in C}(-1)^{x\cdot (c-c_0)}=\sum_{c\in C}(-1)^{x\cdot c}
##

The last equality is because ##C## is a subspace.
 
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martinbn said:
If ##x\in C^\perp## then it is clear. If not, then there is a ##c_0\in C## such that ##x\cdot c_0 =1##. The you have

##
-1\sum_{c\in C}(-1)^{x\cdot c}=(-1)^{x\cdot c}\sum_{c\in C}(-1)^{x\cdot c}=\sum_{c\in C}(-1)^{x\cdot (c-c_0)}=\sum_{c\in C}(-1)^{x\cdot c}
##

The last equality is because ##C## is a subspace.
Thank you very much
 

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