# What is the meaning of the notation $\langle a, b\rangle$ in certain cryptography papers?

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?

• It's inner product. – conchild Jan 21 at 16:13
• See the top of page 9 of that paper. It is the inner product. – Yehuda Lindell Jan 21 at 16:48

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;

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

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