0
$\begingroup$

My question is why the attention head matrices $W^Q$, $W^K$, $W^V$ should not be the same $W = W^Q =W^K= W^V$. In my understanding of transformer-based language models one attention head is responsible for one syntactic or semantic relation between any two words in the context. One might think that such a relation is represented by one matrix $W$ that projects the full word embeddings $x_i$ from their full semantic space to a semantic subspace responsible for this relation. Here we could - in principle - calculate scores $\sigma_{ij}$ as "similiarities" between two projected words $Wx_i$ and $Wx_j$ and then calculate the weighted sum of the projected tokens $Wx_k$.

I wonder why this would not work, and why we need three different matrices.

Another way around: What does it mean to calculate the score as the dot-product of two vectors from two different semantic subspaces? Is this still some kind of similiarity (which lies at the heart of word embeddings)? And doesn't it sound like comparing apples and pears?

Or viewed differently: How similar are the three matrices of an attention head in practice, e.g. when considering some 100$\times$100 attention heads of a large transformer model like ChatGPT?

$\endgroup$
1
  • 1
    $\begingroup$ I found this answer that provides a motivation for using multiple projection matrices. Basically, is also a way to save model parameters. Since the embedding dimension of $x$ is usually larger than the dimensions used for the projections of $Q$, $K$, and $V$. $\endgroup$ Jun 17, 2023 at 14:30

2 Answers 2

1
$\begingroup$

Let me come up with a possibly too naive (and too high-level) answer: Since the weights of the word embedding layer (i.e. the word embeddings) are not trained in advance (via cooccurrence) and independently from the training of the transformer weights (but happens in the same training runs), the word embeddings don't necessarily indicate semantic similarity when taking the dot product (as they do when using Word2Vec or so). So this doesn't have to be expected neither when choosing the same matrix $W = W^Q =W^K= W^V$ in each attention head: the dot product doesn't give scores to be interpreted as similarities but just of some (hard to interpret and only sometimes symmetric) relation.

$\endgroup$
0
$\begingroup$

Actually a multi-head attention (MHA) layer has multiple "heads", where each head is made by three *projection matrices like: $H_i = (W_i^Q, W_i^K, W_i^V)$, for the $i$-th head. A typical number of heads value is eight, for example: each head is responsible of learning to attend to different parts of the sequence, thus you can (partially) interpret transformer-based models by inspecting their attention masks.

A word embedding $x_i$ is not directly fed to MHA. I mean MHA accepts three matrices $Q$ (query), $K$ (key) and $V$ (value), which got projected by $W^Q$, $W^K$ and $W^V$, respectively, when computing the attention scores (i.e. the scaled dot-product attention.) According to the paper (see section 3.2.1) and here, keys $d_k$ and value $d_k$ dimensions can be different, simply because these can be different sequences of (word) embeddings. For example, in a Question-Answer model $Q$ can be the embeddings from the question, while $K$ or $V$ of the expected answer. So learning different weight matrices, $W^Q$, $W^K$, and $W^V$, allows to capture different relevant aspects for each input. Even when $Q=K=V$, i.e. the so called self-attention, the matrices are still fundamental.

In general, consider that queries attend to keys to yield the attention probabilities (via dot-product and softmax) which are used to weight the values. So they basically say which timestep (and so which work embedding in the sequence) is relevant in values (the learned representation by projection.) In other words, the general meaning is which words (in $V$) are relevant to attend at time $t$: so something that goes beyond a simple similarity (which may occur in the self-attention case.)

$\endgroup$
6
  • 1
    $\begingroup$ Thanks, but I think you are wrong in calling the case Q = K = V self-attention. Self-attention means to compare the tokens of one sequence with its own, not of another sequence. Furthermore, my question has nothing to do with multi-head attention, I was asking for the three matrices inside one attention head. $\endgroup$ Jun 15, 2023 at 17:02
  • $\begingroup$ @Hans-PeterStricker I thought you were confusing the three matrices and multi-head attention, sorry. Anyway, the tensorflow's docs say "If $Q$, $K$, $V$ are the same, then this is self-attention." That's what I meant by $Q=K=V$: I should have written that all are the same sequence, indeed. $\endgroup$ Jun 16, 2023 at 15:50
  • $\begingroup$ @Hans-PeterStricker The authors say "instead of perfoming a single attention function [...] we found beneficial to linearly project the queries, keys and values $h$ times with different learned projections [...]" (see beginning of section 3.2.2). So, I think they validated such design with prior experiments. Probably the linear projections "rotate" the representation of $Q$, $K$, and $V$, such that learning is easier. $\endgroup$ Jun 16, 2023 at 15:59
  • 1
    $\begingroup$ Thanks a lot, Luca. To be honest, I find the quote from tensorflow's docs a bit misleading and I dare to doubt that this is a definition of "self-attention". I've always read that self-attention is defined as paying attention of the tokens of a sequence to the tokens of the same sequence. $\endgroup$ Jun 17, 2023 at 7:43
  • $\begingroup$ From tf.keras.layers.Attention: "Inputs are query tensor of shape [batch_size, Tq, dim], value tensor of shape [batch_size, Tv, dim] and key tensor of shape [batch_size, Tv, dim]." It states that Tk = Tv, which doesn't make sense. It must be Tk=Tq, and that's how it's stated in the Attention is all you need paper: "learned linear projections to dk, dk and dv dimensions". So the tensorflow documentation is faulty sometimes, and not too reliable a source? $\endgroup$ Jun 17, 2023 at 7:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .