I'm trying to understand a paper from Brakerski and Vaikuntanathan. In it, they use the notation $\langle a,b \rangle$ but they don't explain what that means.

The answers on this question say that it's just another way to write a tuple, but I don't think that's right in this case since the authors use the regular $(a,b)$ notation in the rest of the paper to denote a tuple.

Does anyone know what else $\langle a,b \rangle$ could mean?

  • 7
    $\begingroup$ It's inner product. $\endgroup$
    – conchild
    Jan 21, 2020 at 16:13
  • 2
    $\begingroup$ See the top of page 9 of that paper. It is the inner product. $\endgroup$ Jan 21, 2020 at 16:48

1 Answer 1


In mathematics, $\langle a,b \rangle$ represents the inner product of two vectors and this is a generalization of the dot product. It is one of the multiplications of the vectors in a Vector space and the result is a scalar.

An inner product satisfies these properties

  1. $\langle c+a,b\rangle =\langle c,a\rangle +\langle a,b\rangle $.

  2. $\langle \alpha a,b\rangle = \alpha\langle a,b\rangle$ where $\alpha$ is a scalar.

  3. $\langle a,b\rangle =\langle b,a\rangle$.

  4. $\langle a,a\rangle = 0 $ and equal if and only if $v=0$.

    • The fourth condition is the positive definite condition. Some authors define the inner product with only the first 3 conditions with a weaker 4th condition $\langle a, b \rangle =0$ for all $b$ then $a=0$.

    As noted in the comments by @levgeni, if the field is $\mathbb{F}_2$ then

    $$\langle (1,1),(1,1) \rangle = 1 \cdot 1 + 1 \cdot 1 = 2 = 0$$ and $$\langle (1,1),(1,0) \rangle = 1 \cdot 1 + 1 \cdot 0 = 1 = 1$$ therefore, the weaker condition is required.

The inner product definition changes according to the vector space. In the context of the paper, let

$$a =(a_1,\ldots,a_n)$$ $$b =(b_1,\ldots,b_n)$$

be two n-dimensional vectors then;

$$\langle a, b \rangle = a_1 \cdot b_1 + \cdots + a_n \cdot b_n $$

  • 1
    $\begingroup$ In a Cryptographic context, the fourth condition is not verified; if the field is $\mathbb{Z}_2$, $\langle \left(1,1\right), \left(1,1\right) \rangle = 1\cdot1 + 1\cdot1 = 0$ $\endgroup$
    – Ievgeni
    Jan 22, 2020 at 13:53
  • $\begingroup$ @levgeni Better now? $\endgroup$
    – kelalaka
    Jan 22, 2020 at 14:24
  • $\begingroup$ I don't see anything to add. It seems okay to me. $\endgroup$
    – Ievgeni
    Jan 22, 2020 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.